Recombination Lines

Spectral lines are narrow ($\Delta \nu \ll \nu$) emission or absorption features in the spectra of gaseous sources. Examples of radio spectral lines include the $\lambda = 21$ cm hyperfine line of interstellar HI, recombination lines of ionized hydrogen and heavier atoms, and rotational lines of polar molecules such as carbon monoxide (CO).

Spectral lines are intrinsically quantum phenomena and have no explanation in classical physics. The classical concepts of ideal particles and waves are blurred in quantum mechanics: classical particles (e.g., electrons) and waves (e.g., radio waves) have the characteristics of both particles and waves. The quantum of action is Planck's constant $h\approx 6.62607 \times 10^{-27}$ erg s; its dimensions are (mass$\times$length$^2 \times$time$^{-1}$), the same as (energy$\times$time) or (angular momentum) or (length$\times$momentum). Particles of momentum $p$ have De Broglie wavelengths $\lambda = h / p$. Radiation is quantized into photons whose energy is proportional to frequency: $E = h \nu$. Spectral lines have definite frequencies resulting from transitions between discrete energy states in physical systems, and these discrete states often arise from quantization of angular momentum. Another quantum effect important to spectral lines, particularly at radio wavelengths where $h \nu \ll k T$, is stimulated emission. Fortunately, the fundamental characteristics of radio spectral lines can be derived from fairly simple and basic quantum mechanics and thermodynamics.

Spectral lines are powerful diagnostics of conditions in astronomical objects. Doppler shifts of line frequencies measure radial velocities. These velocities yield the redshifts and Hubble distances of extragalactic sources, as well as the rotation curves and radial mass distributions for resolved galaxies. Collapse speeds, turbulent velocities, and thermal motions contribute to line broadening in Galactic sources. The physical conditions (e.g., temperature and density) and chemistry of HII regions, dust-obscured dense molecular clouds, and diffuse interstellar gas are constrained by spectral-line observations. Some characteristics of radio spectral lines include:

(1) Natural line widths much smaller than Doppler-broadened line widths, so very small changes in radial velocity can be measured.

(2) $h \nu \ll k T$ so stimulated emission is important. This causes line opacities to vary as $T^{-1}$ and favors the formation of natural masers.

(3) The ability to penetrate dust in our Galaxy and in other galaxies, allowing the detection of line emission from dusty molecular clouds, protostars, and molecular disks orbiting AGNs.

(4) In practice, frequency (inverse time) can be measured with much higher precision than wavelength (length), so very sensitive searches for small changes in the fundamental physical constants over cosmic timescales can be made.

Most of the interstellar medium (ISM) in our Galaxy is in rough pressure equilibrium because mass motions with speeds up to the speed of sound act to reduce pressure gradients quickly. Temperatures equilibrate more slowly, so there are wide ranges of the (temperature$\times$density) product consistent with a given pressure. Consequently, there are four important phases of the ISM having comparable pressures:

(1) cold (10's of K) dense molecular clouds

(2) cool ($\sim 10^2$ K) neutral HI gas

(3) warm ($\sim 10^4$ K) ionized HII gas

(4) hot ($\sim 10^6$ K) low-density ionized gas (in bubbles formed by expanding supernova remnants, for example).

All but the hot phase are sources of radio spectral lines.

Recombination Lines of Excited Atoms

The semiclassical Bohr atom contains a nucleus of protons and neutrons around which electrons move in circular orbits.  The nuclear mass $M$ is always much greater than the electron mass $m_{\rm e}$. The electrons have De Broglie wavelengths
$$\lambda = {h \over p} = { h \over m_{\rm e} v}~,$$
where $v$ is the electron speed.  Only those orbits whose circumferences equal an integer number $n$ of wavelengths are permitted. Thus the Bohr radius $a_{\rm n}$ of the $n$th permitted electron orbit satisfies the quantization rule
$$ 2 \pi a_{\rm n} = n \lambda = { n h \over m_{\rm e} v}~.$$
The requirement that
$$a_{\rm n} = { n h \over 2 \pi m_{\rm e} v}
= {n \hbar \over m_{\rm e} v}$$
is equivalent to the requirement that the orbital angular momentum $L = m_{\rm e} v a_{\rm n}$ be an integer multiple of $\hbar \equiv h / (2 \pi)$.

The relation between $a_{\rm n}$ and $v$ is provided by balancing the Coulomb and centrifugal forces. For a hydrogen atom,
$$ {e^2 \over a_{\rm n}^2} = {m_{\rm e} v^2 \over a_{\rm n}}$$
so
$$v^2 = {e^2 \over m_{\rm e} a_{\rm n}}$$
$$\bbox[border:3px blue solid,7pt]{a_{\rm n} = {n^2 \hbar^2 \over m_{\rm e} e^2}}\rlap{\quad \rm {(7A1)}}$$

Example: What is the Bohr radius of a hydrogen atom whose electron is in the $n$th electronic energy level?
$$a_{\rm n} = { \hbar^2 \over m_{\rm e} e^2} n^2 = {[6.63/(2 \pi) \times 10^{-27} {\rm ~erg~s}]^2 \over 9.11 \times 10^{-28} {\rm ~g} \times (4.8 \times 10^{-10} {\rm ~statcoul})^2} n^2 \approx 0.53 \times 10^{-8} {\rm ~cm} \times n^2$$
The Bohr radius of a hydrogen atom in its ground electronic state ($n = 1$) is only $a_1 \approx 0.53 \times 10^{-8}$ cm, but a highly excited ($n \approx 100$) radio-emitting atom in the ISM can be remarkably large: $a_{100} \approx 0.53 \times 10^{-4}$ cm.

The electron in a Bohr atom can fall from the level $(n + \Delta n)$ to $n$, where $\Delta n$ and $n$ are any natural numbers ($1,~2,~3, ...$) by emitting a photon whose energy equals the energy difference $\Delta E$ between the initial and final levels.  Astronomers label a recombination line by the name of the element, the final level number $n$, and a Greek letter denoting $\Delta n$: $\alpha$ for $\Delta n = 1$, $\beta$ for $\Delta n = 2$, $\gamma$ for $\Delta n = 3$, etc.  For example, the line produced by the transition from $n = 92$ to $n = 91$ of a hydrogen atom is called the H91$\alpha$ line. 

The total electronic energy $E_{\rm n}$ is the sum of the kinetic and potential energies of the electron in the $n$th level:
$$E_{\rm n} = T + V = -T = V / 2 = -{e^2 \over 2 a_{\rm n}} = -{e^2 m_{\rm e} e^2 \over 2 n^2 \hbar^2} = - {m_{\rm e} e^4 \over 2 \hbar^2 n^2}$$
The electronic energy change $\Delta E$ going from level $(n + \Delta n)$ to level $n$ is equal to the energy $h \nu$ of the emitted photon:
$$\Delta E = {m_{\rm e} e^4 \over 2 \hbar^2} \biggl[ {1 \over n^2} - {1 \over (n + \Delta n)^2} \biggr] = h \nu~,$$
so the photon frequency is
$$\nu = \biggl( { 2 \pi^2 m_{\rm e} e^4 \over h^3 c} \biggr) c \biggl[ {1 \over n^2} - {1 \over (n + \Delta n)^2} \biggr]$$
The quantity in parenthesis is called the Rydberg constant $R_\infty$, where the subscript refers to our assumption that the nuclear mass $M$ is essentially infinite.
$$R_\infty \equiv \biggl({2 \pi^2 m_{\rm e} e^4 \over h^3 c}\biggr) \approx {2 \pi^2 \cdot 9.11 \times 10^{-28} {\rm ~g~} \cdot (4.8 \times 10^{-10} {\rm ~esu})^4 \over (6.63 \times 10^{-27} {\rm ~erg~s})^3 \cdot 3 \times 10^{10} {\rm ~cm~s}^{-1}}$$
$$R_\infty = 1.09737312\dots \times 10^5 {\rm ~cm}^{-1}$$
The dimensions of $R_\infty$ are length$^{-1}$, and the product $R_\infty c$ is the Rydberg frequency $\nu \approx 3.28984 \times 10^{15}$ Hz.

If we allow for the finite nuclear mass $M$ and repeat the above analysis in the atomic center-of-mass frame, we get the same frequency formula with $R_\infty$ replaced by $R_{\rm M}$:
$$\bbox[border:3px blue solid,7pt]{\nu = R_{\rm M} c \biggl[ { 1 \over n^2} - {1 \over (n + \Delta n)^2} \biggr] \qquad {\rm where} \qquad R_{\rm M} \equiv R_\infty \biggl( 1 + {m_{\rm e} \over M} \biggr)^{-1}}\rlap{\quad \rm {(7A2)}}$$
The hydrogen nucleus is a single proton of mass $m_{\rm p} \approx 1836.1 m_{\rm e}$, so $M({\rm H}) \approx 1836.1 m_{\rm e}$. 

Example: The H109$\alpha$ transition from ($n + \Delta n = 110$) to $n = 109$ has $\Delta n = 1$ and frequency
$$\nu = R_{\rm M} c \biggl[ { 1 \over n^2} - {1 \over (n + \Delta n)^2} \biggr] \qquad {\rm where} \qquad R_{\rm M} \equiv R_\infty \biggl( 1 + {m_{\rm e} \over M} \biggr)^{-1} $$
$$R_{\rm M} c = 3.28984 \times 10^{15}\,{\rm Hz}
\biggl( 1 + {1 \over 1836.1} \biggr)^{-1} = 3.28805 \times 10^{15}\,{\rm Hz}$$
$$\nu = 3.28805 \times 10^{15}\,{\rm Hz}
\biggl( {1 \over 109^2} - {1 \over 110^2} \biggr) = 5.0089 \times 10^9\,{\rm Hz}$$

The mass of a neutron is about equal to the mass of a proton so the $^4$He nucleus consisting of two protons and two neutrons has mass $M(^4{\rm He}) \approx 4 M({\rm H})$, the isotope of carbon with six protons and six neutrons has $M(^{12}{\rm C}) \approx 12 M({\rm H})$, and so on.  Electrons recombining onto singly ionized atoms with any number $N_{\rm p}$ of protons and $N_{\rm p}-1$ electrons see a net charge of one proton, so the recombination-line spectra of heavier atoms are very similar to that of hydrogen, but the lines of heavier atoms are at slightly higher frequencies and may be detected individually. For example, the primordial abundance of the rare helium isotope $^3$He is important because it reflects the density of baryons in the early universe. The abundance of $^3$He in galactic HII regions has been measured via radio recombination line emission and indicates that baryons account for only a few percent of the critical density needed to close the universe.

Thus a simpler (but not very accurate) approximation for the radio frequency $\nu$ is
$$\nu \approx {2 (R_{\rm M} c) \Delta n \over n^3}$$
and the frequency separation $\Delta \nu = \nu(n) - \nu(n+1)$ between adjacent lines can be written
$$\bbox[border:3px blue solid,7pt]{{\Delta\nu \over \nu} \approx {3 \over n}}\rlap{\quad \rm {(7A3)}}$$

The H109$\alpha$ line was first radio detected by P. Mezger in 1965, despite (incorrect) theoretical predictions that pressure broadening would smear out the lines and make them undetectable.  It is true that atomic collisions in the interstellar medium significantly disturb the energy levels of large atoms, but this disturbance is about the same for adjencent energy levels, so the differential disturbance that affects the line frequency is actually much smaller.  His conclusion: never don't make an observation just because you have been told that it won't work.

Next we consider the spontaneous emission rate—how quickly does an isolated atom with $n \gg 1$ decay to a lower energy level? A rigorous answer requires the solution of a complex problem in quantum mechanics. However, we can get a fairly good answer by noting that most photons are emitted by $\Delta n = 1$ transitions and, by the correspondence principle, the time-averaged radiated power is given by the classical Larmor's equation for a dipole with dipole moment $e a_{\rm n}$. 
$$\langle P \rangle= {2 e^2 \over 3 c^3} (\omega^2 a_{\rm n})^2
\langle \cos^2(\omega t) \rangle .$$
$$\langle P \rangle = {2 e^2 \over 3 c^3} (2 \pi \nu)^4 {a_{\rm n}^2 \over 2}$$
The photon emission rate (s$^{-1}$) is this average power emitted by one atom divided by the energy of the emitted photon. This rate is called the spontaneous emission rate. The spontaneous emission rate from transitions from level $n$ to level $(n-1)$ is denoted by $A_{{\rm n}, {\rm n}-1}$.
$$A_{{\rm n}, {\rm n}-1} = {\langle P \rangle \over h \nu}~,$$
where
$$\nu \approx {2 R_\infty c \Delta n \over n^3}$$
in the limit $n \gg \Delta n$.  Also in that limit, $A_{{\rm n+1},{\rm n}} \approx A_{{\rm n}, {\rm n}-1}$.

Recall that
$$\nu \approx {2 R_\infty c \over n^3} {\rm \qquad and \qquad} a_{\rm n} \approx {n^2 h^2 \over 4 \pi^2 m_{\rm e} e^2}$$
so
$$A_{{\rm n+1}, {\rm n}} \approx {\langle P \rangle \over h \nu} \approx {2 e^2 \over 3 c^3} \biggl({16 \pi^4 \nu^3 \over h}\biggr) {a_{\rm n}^2 \over 2}$$

$$A_{{\rm n+1}, {\rm n}} \approx {16 \pi^4 \over 3}{e^2 \over c^3 h} \biggl({2 R_\infty c \over n^3}\biggr)^3 \biggl({n^2 h^2 \over 4 \pi^2 m_{\rm e} e^2}\biggr)^2$$

$$A_{{\rm n+1}, {\rm n}} \approx {16 \pi^4 \over 3} {e^2 \over c^3 h} \biggl({4 \pi^2 m_{\rm e} e^4 \over h^3}\biggr)^3 \biggl({h^2 \over 4 \pi^2 m_{\rm e} e^2}\biggr)^2 {1 \over n^5}$$

$$\bbox[border:3px blue solid,7pt]{A_{{\rm n+1},{\rm n}} \approx {64 \pi^6 m_{\rm e} e^{10} \over 3  c^3 h^6 n^5}}\rlap{\quad \rm {(7A4)}}$$

Evaluating the constants gives
$$A_{{\rm n+1},{\rm n}} \approx \biggl[ { 64 \pi^6 \cdot 9.11 \times 10^{-28} {\rm ~g} \cdot (4.8 \times 10^{-10} {\rm ~statcoul})^{10} \over 3 \cdot (3 \times 10^{10} {\rm ~cm~s}^{-1})^3  \cdot(6.63 \times 10^{-27} {\rm ~erg~s})^6} \biggr] {1 \over n^5}$$
$$\bbox[border:3px blue solid,7pt]{A_{{\rm n+1}, {\rm n}} \approx 5.3 \times 10^{9} \biggl({1 \over n^5}\biggr) {\rm ~s}^{-1}}\rlap{\quad \rm {(7A5)}}$$

Example: The 5.08 GHz H109$\alpha$ transition rate is $A_{110,109} \approx 0.3$ s$^{-1}$.

The associated natural or intrinsic line width from the uncertainty principle is
$$\Delta \nu \sim A_{{\rm n+1}, {\rm n}}^{-1} / \pi \sim 1 {\rm ~Hz}$$
It is negligibly small at the large $n$ that produce radio-frequency photons. Thus the actual line profile (normalized intensity as a function of frequency) is determined by Doppler shifts indicating radial velocities $v_{\rm r}$. These motions may be microscopic (thermal) or macroscopic (indicating large-scale turbulence, flows, or rotation). In the nonrelativistic limit $v_{\rm r} \ll c$, the Doppler equation relating the observed frequency $\nu$ to the line rest frequency $\nu_0$ is
$$\nu \approx \nu_0 \biggl( 1 - {v_{\rm r} \over c} \biggr)$$
so the radial velocity can be estimated from
$$v_{\rm r} \approx {c ( \nu_0 - \nu) \over \nu_0}~.$$

The thermal component of the line profile can be calculated from the Maxwellian speed distribution of atoms with mass $M$ and temperature $T$. The speed in any one coordinate of an isotropic distribution is $3^{-1/2}$ lower than the total speed in three dimensions so
$$f(v_{\rm r}) = \biggl( {M \over 2 \pi k T} \biggr)^{1/2} \exp \biggl( - {M v_{\rm r}^2 \over 2 k T} \biggr)$$
is the normalized ($\int f(v_{\rm r}) d v_{\rm r} = 1$) radial velocity distribution. The normalized line profile $\phi(\nu)$ for thermal emission is
$$\vert \phi(\nu) d \nu \vert = f(v_{\rm r}) d v_{\rm r}$$
$$\phi(\nu) = \biggl( { M \over 2 \pi k T} \biggr)^{1/2} \exp \biggl[ - {M \over 2 k T} {c^2 ( \nu - \nu_0)^2 \over \nu_0^2 } \biggr] \bigg| {d v_{\rm r} \over d \nu } \bigg|$$
$$\bbox[border:3px blue solid,7pt]{\phi(\nu) = {c \over \nu_0} \biggl( { M \over 2 \pi k T} \biggr)^{1/2} \exp \biggl[ - {M c^2 \over 2 k T} {(\nu - \nu_0)^2 \over \nu_0^2 } \biggr]}\rlap{\quad \rm {(7A6)}}$$

This is a Gaussian line profile. Its full width between half-maximum points (FWHM) $\Delta \nu$ is the solution of $$\exp \biggl[ - {M c^2 \over 2 k T} { (\Delta \nu / 2)^2 \over \nu_0^2} \biggr] = {1 \over 2}$$ $${Mc^2 \over 2 k T} {\Delta \nu^2 \over 4 \nu_0^2} = \ln 2$$

$$\bbox[border:3px blue solid,7pt]{\Delta \nu = \biggl( { 8 \ln 2 \, k \over c^2} \biggr)^{1/2} \biggl( { T \over M} \biggr)^{1/2} \nu_0}\rlap{\quad \rm {(7A7)}}$$

Example: What is the FWHM of the H109$\alpha$ line ($\nu_0 = 5.08$ GHz) in a quiescent (no macroscopic motions) HII region with temperature $T \approx 10^4$ K?

$$\Delta \nu \approx \biggl[ { 8 \ln 2\, \cdot \, 1.38 \times 10^{-16} {\rm ~erg~K}^{-1} \over ( 3 \times 10^{10} {\rm ~cm~s}^{-1})^2 } \biggr]^{1/2} \biggl( { 10^4 {\rm ~K} \over 1836 \cdot 9.11 \times 10^{-28} {\rm ~g} } \biggr)^{1/2}\cdot5.08\times10^9{\rm ~Hz}$$ $$\Delta \nu = 3.6 \times 10^5 {\rm ~Hz}$$

Normalization (requiring $\int \phi(\nu) d \nu = 1$) implies that the value of $\phi$ at the line center ($\nu = \nu_0$) is
$$\phi(\nu_0) = {c \over \nu_0} \biggl( { M \over 2 \pi k T} \biggr)^{1/2}$$ $$\phi(\nu_0) = {c \over \Delta \nu} \biggl( { 8 \ln 2 \, k T \over M c^2} {M \over 2 \pi k T} \biggr)^{1/2}$$

$$\bbox[border:3px blue solid,7pt]{\phi(\nu_0) = \biggl( { \ln 2 \over \pi} \biggr)^{1/2} {2 \over \Delta \nu}}\rlap{\quad \rm {(7A8)}}$$