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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}}

\usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document} \def\lya{Ly\alpha} \subsection*{ Einstein A's for $\lya$}

$\ato$ is a measure of the probability of decay per unit time, so $\ato^{-1}\sim lifetime$. This should be about equal to the energy of the electron state divided by the average power radiated by an electron being accelerated: $$\ato^{-1}\sim{E\over P}\sim{\hbar\omega_0\over{2\over3}{e^2\ddot x^2 \over c^3}} \sim{3\hbar\omega_0c^3\over2(e\ddot x)^2}$$ Now $e\cdot\vec x=\vec d$ (the electric dipole moment) and $\ddot x\sim\omega_0^2x$ for a spring, so: $$\ato^{-1}\sim{3\hbar\wz c^3\over2d^2\wz^4}$$ $$\boxed{\ato\sim{2d^2\wz^3\over3\hbar c^3}}$$ For H: $d\sim ea_0$ and $\lambda_\lya=1216\AA$ so: $$\ato\sim5\e8s^{-1}$$

\subsection*{ Magnetic Dipole for $\lya$}

The magnetic dipole of an electron is: $$\mu_e={e\hbar\over m_ec}$$ Thus we can estimate the ratio of $\ato$ for magnetic dipole transitions to that of electric dipole transitions: $${\ato\eval{mag}\over\ato\eval{elec}}\sim\left({\mu_e\over d}\right)^2 \sim\left({e^2\over\hbar c}\right)^2\sim\alpha^2$$ This tells us that the magnetic dipole states (that is, fine and hyperfine states) are longer lived than electric dipole states by a factor of $\alpha^2$. $${\ato\eval{21cm}\over\ato\eval{Ly\alpha}}\sim\alpha^2 \left({1216\AA\over21cm}\right)^3$$ $$\ato\eval{21cm}\sim6\e{-15}s^{-1}$$ The actual value is $2.876\e{-15}s^{-1}$.

If one is nearby a rotating quadrupole, one sees the $\mfe$ (electric) field rotating rigidly. However, from far away, there are kinks in the field, resulting in a retarded potential. The radiation nearby goes as $r_{near}\sim\lambda$. For a monopole, the electric field is $\mfe\sim{q\over r^2}$. For a dipole, it is $\mfe={q\over r^2} {s\over r}$, where s is the charge separation. For a quadrupole: $$\mfe={q\over r^2}\left({s\over r}\right)^2$$ Since $P\propto \mfe^2$, the ratio of the powers emitted by a quadrupole vs. a dipole should be: $${P_{quad}\over P_{di}}\sim\left({s\over r}\right)^2 \sim\left({s\over\lambda}\right)^2$$ An acoustic analogy: a kettle whistle is a monopole, a guitar string is a dipole, and a tuning fork (with its two out-of-phase prongs) is a quadrupole.\\

Anyway, since $\ato\sim{P\over E}$, $$\boxed{{\ato\eval{quad}\over\ato\eval{di}} \sim\left({s\over\lambda}\right)^2}$$ Thus $28\mu m$, the lowest quadrupole rotational transition of $H_2$, should have an $\ato$ of about: $$\ato\eval{28\mu m}\sim\ato\eval\lya\left({s\over\lambda_{H_2}}\right)^2 \left({\lambda_\lya\over \lambda_{H_2}}\right) \sim\ato\eval\lya\left({a_0\over 28\mu m}\right)^2 \left({1216\AA\over 28\mu m}\right)^3 \sim7\e{-11}s^{-1}$$ The actual value is $3\e{-11}s^{-1}$.

In HI, the $n=110\to n=109$ transition has a wavelength of 6 cm. We can estimate its $\ato$: $$\ato\eval{6cm}\sim\ato\eval{Ly\alpha} \underbrace{\left({1216\AA\over6cm}\right)^3}_{change\atop in\ \lambda} \underbrace{\left({a_{110}\over a_0}\right)^2}_{change\atop in\ atom\ size}$$ \def\bra#1{\langle #1|} \def\ket#1{|#1\rangle} This presents the question of which dipole moment to use. It turns out we must use $\bra{i}\vec k\cdot\vec r\ket{f}$.
\subsection*{ Back to $\sigma$}
$$\sigot\eval{line\atop center}\sim{\lambda^2\over8\pi}{\ato\over\Delta\nu}$$ Now $\Delta\nu\sim\nu$ for doppler broadening, and $\ato\sim\nu^3$, so for electric and magnetic dipole transitions: $$\sigot\sim\lambda^0$$ So the cross-section for these transitions does not depend on wavelength.
In general, we will be talking about a plane wave of wavelength $\lambda$ which is incident upon a particle (grain) of radius a. The cross-section for absorption, scattering, and emission will all be proportional to the physical cross-section of the grain, with some {\it absorption efficiency} Q: $$\sigma_i=Q_i\pi a^2$$ Kirchkoff's Law requires that $Q_{emit}=Q_{abs}$. If $a\gg\lambda$, then \def\qscat{Q_{scat}} \def\qabs{Q_{abs}} we have the geometric optics limit, and $\qscat+\qabs\sim$. In fact, Babinet's Principle says: $$\boxed{\qscat+\qabs=2}$$ The proof of this goes as follows: suppose we have an infinite plane wave focused by an infinite lens onto a point on a wall. The power pattern should be a delta function at that point on the wall. Now suppose you place a screen with an aperture of diameter $a$ between the plane wave and the lens. We should now see an interference pattern on the wall. Call the power incident on a point on the wall $P_1$. Now put in a new aperture which is the exact compliment of our previous one (it is a scattering body of diameter $a$), we should see a new power at our point on the wall: $P_2$. However, the sum of waves incident on the wall under these two apertures should be the same as the wave from the sum of the apertures, which was a delta function. Thus: $$P_1=P_2$$ The first aperture (the slit) represented the scattering power from diffraction alone $(\qscat=1)$, and the second represented the absorbed power. The sum of these two powers, for everywhere but the true focus point, is actually twice the incident power, and since $P_1=P_2$, we get: $$\qscat+\qabs=2$$ Note that if $\qscat>1$, then the object in shiny: more light gets diffracted and less gets absorbed. Thus, although $\qscat$ changes, the sum remains the same.