# Estimating the Strength of Atomic/Molecular Lines

## 1 Einstein A’s for ${\displaystyle {Ly\alpha }}$

Recall from Einstein Coefficients that ${\displaystyle {A_{21}}}$ is a measure of the probability of decay per unit time, so ${\displaystyle {A_{21}}^{-1}}$ is approximately the lifetime of an atom in an excited state. This should be about equal to the energy of the electron state divided by the average power radiated by an electron being accelerated:

${\displaystyle {A_{21}}^{-1}\sim {E \over P}\sim {\hbar \omega _{0} \over {2 \over 3}{e^{2}{\ddot {x}}^{2} \over c^{3}}}\sim {3\hbar \omega _{0}c^{3} \over 2(e{\ddot {x}})^{2}}\,\!}$

Now ${\displaystyle e\cdot {\vec {x}}={\vec {d}}}$ (the electric dipole moment) and ${\displaystyle {\ddot {x}}\sim \omega _{0}^{2}x}$ for a spring, so:

${\displaystyle {A_{21}}^{-1}\sim {3\hbar \omega _{0}c^{3} \over 2d^{2}\omega _{0}^{4}}\,\!}$
${\displaystyle {{A_{21}}\sim {2d^{2}\omega _{0}^{3} \over 3\hbar c^{3}}}\,\!}$

For H: ${\displaystyle d\sim ea_{0}}$ and ${\displaystyle \lambda _{Ly\alpha }=1216\mathrm {\AA} }$ so:

${\displaystyle {A_{21}}\sim 5\cdot 10^{8}s^{-1}\,\!}$

## 2 Magnetic Dipole for ${\displaystyle {Ly\alpha }}$

The magnetic dipole of an electron is:

${\displaystyle \mu _{e}={e\hbar \over m_{e}c}\,\!}$

Thus we can estimate the ratio of ${\displaystyle {A_{21}}}$ for magnetic dipole transitions to that of electric dipole transitions:

${\displaystyle {{A_{21}}{\big |}_{mag} \over {A_{21}}{\big |}_{elec}}\sim \left({\mu _{e} \over d}\right)^{2}\sim \left({e^{2} \over \hbar c}\right)^{2}\sim \alpha ^{2}\,\!}$

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 ${\displaystyle \alpha ^{2}}$.

${\displaystyle {{A_{21}}{\big |}_{21cm} \over {A_{21}}{\big |}_{Ly\alpha }}\sim \alpha ^{2}\left({1216\mathrm {\AA} \over 21cm}\right)^{3}\,\!}$
${\displaystyle {A_{21}}{\big |}_{21cm}\sim 6\cdot 10^{-15}s^{-1}\,\!}$

The actual value is ${\displaystyle 2.876\cdot 10^{-15}s^{-1}}$.

If one is nearby a rotating quadrupole, one sees the ${\displaystyle {\mathfrak {E}}}$ (electric) field rotating rigidly. However, from far away, there are kinks in the field, resulting in a retarded potential. The radiation nearby goes as ${\displaystyle r_{near}\sim \lambda }$. For a monopole, the electric field is ${\displaystyle {\mathfrak {E}}\sim {q \over r^{2}}}$. For a dipole, it is ${\displaystyle {\mathfrak {E}}={q \over r^{2}}{s \over r}}$, where s is the charge separation. For a quadrupole:

${\displaystyle {\mathfrak {E}}={q \over r^{2}}\left({s \over r}\right)^{2}\,\!}$

Since ${\displaystyle P\propto {\mathfrak {E}}^{2}}$, the ratio of the powers emitted by a quadrupole vs. a dipole should be:

${\displaystyle {P_{quad} \over P_{di}}\sim \left({s \over r}\right)^{2}\sim \left({s \over \lambda }\right)^{2}\,\!}$

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 ${\displaystyle {A_{21}}\sim {P \over E}}$,

${\displaystyle {{{A_{21}}{\big |}_{quad} \over {A_{21}}{\big |}_{di}}\sim \left({s \over \lambda }\right)^{2}}\,\!}$

Thus ${\displaystyle 28\mu m}$, the lowest quadrupole rotational transition of ${\displaystyle H_{2}}$, should have an ${\displaystyle {A_{21}}}$ of about:

${\displaystyle {A_{21}}{\big |}_{28\mu m}\sim {A_{21}}{\big |}_{Ly\alpha }\left({s \over \lambda _{H_{2}}}\right)^{2}\left({\lambda _{Ly\alpha } \over \lambda _{H_{2}}}\right)^{3}\sim {A_{21}}{\big |}_{Ly\alpha }\left({a_{0} \over 28\mu m}\right)^{2}\left({1216\mathrm {\AA} \over 28\mu m}\right)^{3}\sim 7\cdot 10^{-11}s^{-1}\,\!}$

The actual value is ${\displaystyle 3\cdot 10^{-11}s^{-1}}$.

In HI, the ${\displaystyle n=110\to n=109}$ transition has a wavelength of 6 cm. We can estimate its ${\displaystyle {A_{21}}}$:

${\displaystyle {A_{21}}{\big |}_{6cm}\sim {A_{21}}{\big |}_{Ly\alpha }\underbrace {\left({1216\mathrm {\AA} \over 6cm}\right)^{3}} _{change \atop in\ \lambda }\underbrace {\left({a_{110} \over a_{0}}\right)^{2}} _{change \atop in\ atom\ size}\,\!}$

This presents the question of which dipole moment to use. It turns out we must use ${\displaystyle \langle i|{\vec {k}}\cdot {\vec {r}}|f\rangle }$.

## 5 Back to ${\displaystyle \sigma }$

${\displaystyle \sigma _{12}{\big |}_{line \atop center}\sim {\lambda ^{2} \over 8\pi }{{A_{21}} \over \Delta \nu }\,\!}$

Now ${\displaystyle \Delta \nu \sim \nu }$ for doppler broadening, and ${\displaystyle {A_{21}}\sim \nu ^{3}}$, so for electric and magnetic dipole transitions:

${\displaystyle \sigma _{12}\sim \lambda ^{0}\,\!}$

So the cross-section for these transitions does not depend on wavelength.