# Estimating the Strength of Atomic/Molecular Lines

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

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

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

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 ${\displaystyle \lambda _{L}y\alpha =1216\mathrm {\AA} }$ 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:

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

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.