Recombination Lines

Introduction to Radio Spectral 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. Classical particles and waves are idealized concepts like infinitesimal points or perfectly straight lines in geometry; they don't exist in the real world.  Some things are nearly waves (e.g., radio waves) and others are nearly particles (e.g., electrons), but all share characteristics of both particles and waves. Unlike ideal waves, real radio waves do not have a continuum of possible energies.  Instead, electromagnetic radiation is quantized into photons whose energy is proportional to frequency: $E = h \nu$.  Unlike ideal particles, real particles of momentum $p$ have wave functions whose De Broglie wavelength is $\lambda = h / p$.  An electron's orbit about the nucleus of an atom must allow standing waves, so its circumference must be an integer number of wavelengths. The Planck's constant $h\approx 6.62607 \times 10^{-27}$ erg s in these two equations is a quantum of action; its dimensions are (mass$\times$length$^2 \times$time$^{-1}$), the same as (energy$\times$time) or (angular momentum) or (length$\times$momentum). Spectral lines have definite frequencies resulting from transitions between discrete energy states in physical systems, and these discrete states 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 from interstellar atoms and molecules can be derived from fairly simple applications of quantum mechanics and thermodynamics.

Spectral lines are powerful diagnostics of physical and chemical 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. Temperatures, densities, and chemical compositions of HII regions, dust-obscured dense molecular clouds, and diffuse interstellar gas are constrained by spectral-line data. Some characteristics of radio spectral lines include:

(1) The "natural" line widths are much smaller than Doppler-broadened line widths, so very small changes in radial velocity can be measured.
(2) Stimulated emission is important because $h \nu \ll k T$. 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 allows the detection of line emission emerging 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 Line Frequencies

The semiclassical Bohr atom contains a nucleus of protons and neutrons around which one or more electrons move in circular orbits.  The nuclear mass $M$ is always much greater than the sum of the electron masses $m_{\rm e}$, so the nucleus is nearly at rest in the center-of-mass frame. The wave functions of the electrons have De Broglie wavelengths
$$\lambda = {h \over p} = { h \over m_{\rm e} v}~,$$ where $p$ is the electron's momentum and $v$ is its speed.  Only those orbits whose circumferences equal an integer number $n$ of wavelengths correspond to standing waves and 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
$$a_{\rm n} = { n h \over 2 \pi m_{\rm e} v}
= {n \hbar \over m_{\rm e} v}$$ implies 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 since $a_{\rm n} \propto n^2$ a highly excited ($n \approx 100$) radio-emitting atom in the ISM can be remarkably large: $2a_{100} \approx 10^{-4}$ cm $= 1 \mu$m, which is bigger than most viruses!

The radius of the nth Bohr orbit is proportional to $n^2$, so radio-emitting hydrogen atoms with $n \sim 100$ are much bigger than ordinary hydrogen atoms in the $n=1$ ground state.

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.  Such spectral lines are called recombination lines because formerly free electrons recombining with ions quickly cascade to the ground state by emitting photons. Astronomers label each recombination line by the name of the element, the final level number $n$, and successive letters in the Greek alphabet to denote the level change $\Delta n$: $\alpha$ for $\Delta n = 1$, $\beta$ for $\Delta n = 2$, $\gamma$ for $\Delta n = 3$, etc.  For example, the recombination line produced by the transition between the $n = 92$ and $n = 91$ levels 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 circular orbit:
$$E_{\rm n} = T + V = -T = V / 2 = -{e^2 \over 2 a_{\rm n}} = -{e^2 m_{\rm e} e^2 \over 2 n^2 \hbar^2} = - {m_{\rm e} e^4 \over 2 \hbar^2 n^2}$$ 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 factor in large parentheses is called the Rydberg constant $R_\infty$, where the subscript refers to the limit of infinite nuclear mass $M$.
$$R_\infty \equiv \biggl({2 \pi^2 m_{\rm e} e^4 \over h^3 c}\biggr) \approx {2 \pi^2 \cdot 9.11 \times 10^{-28} {\rm ~g~} \cdot (4.8 \times 10^{-10} {\rm ~esu})^4 \over (6.63 \times 10^{-27} {\rm ~erg~s})^3 \cdot 3 \times 10^{10} {\rm ~cm~s}^{-1}}$$ $$R_\infty = 1.09737312\dots \times 10^5 {\rm ~cm}^{-1}$$ 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.

Allowing for the finite nuclear mass $M$ and repeating the analysis above in the atomic center-of-mass frame yields 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: What is the frequency of the photon produced by the H109$\alpha$ transition from $n + \Delta n = 110$ to $n = 109$?
$$\nu = R_{\rm M} c \biggl[ { 1 \over n^2} - {1 \over (n + \Delta n)^2} \biggr] \qquad {\rm where} \qquad R_{\rm M} \equiv R_\infty \biggl( 1 + {m_{\rm e} \over M} \biggr)^{-1} $$ $$R_{\rm M} c = 3.28984 \times 10^{15}\,{\rm Hz}
\biggl( 1 + {1 \over 1836.1} \biggr)^{-1} = 3.28805 \times 10^{15}\,{\rm Hz}$$ $$\nu = 3.28805 \times 10^{15}\,{\rm Hz}
\biggl( {1 \over 109^2} - {1 \over 110^2} \biggr) \approx 5.0089 \times 10^9\,{\rm Hz}$$

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 orbit in the potential produced by 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 (Eq. 7A2) 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.

radio recombination line data

Observed recombination-line spectra from the 91$\alpha$ and 92$\alpha$ transitions of hydrogen, helium, and carbon observed in an HII region (Quireza et al. 2006, ApJS, 165, 338).

The stronger radio recombination lines are produced by transitions with $\Delta n \ll n$, so we can sometimes use the approximation
$$\biggl[ { 1 \over n^2} - { 1 \over (n + \Delta n)^2} \biggr] \approx { (n + \Delta n)^2 - n^2 \over n^2 (n + \Delta n)^2} = {n^2 + 2 n \Delta n + (\Delta n)^2 - n^2 \over n^2 [n^2 + 2 n \Delta n + (\Delta n)^2]} \approx {2 n \Delta n \over n^4} = {2 \Delta n \over n^3}$$ Thus a simpler (but not extremely 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 is about $$\bbox[border:3px blue solid,7pt]{{\Delta\nu \over \nu} \approx {3 \over n}}\rlap{\quad \rm {(7A3)}}$$ Adjacent high-$n$ (low $\nu$) radio recombination lines have small fractional frequency separations, so two or more lines can often be observed simultaneously.

The $\Delta n = 1$ hydrogen recombination lines, shown here as vertical bars, are closely spaced in frequency.

The H109$\alpha$ line was first detected by P. Mezger in 1965, despite (incorrect) theoretical predictions that pressure broadening would smear out the lines in frequency 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 adjacent energy levels, so the differential disturbance that alters the line frequency is actually much smaller.  His conclusion: Don't abandon an observation just because you have been told that it won't work.

Recombination Line Strengths

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 quantum mechanics. However, we can get a fairly good answer by noting that radio photons are emitted by atoms with $n \gg 1$ and use the correspondence principle, Bohr's hypothesis that systems with large quantum numbers behave almost classically.  The time-averaged radiated power for $n \gg 1$ transitions 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 for 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
$$a_{\rm n} \approx {n^2 h^2 \over 4 \pi^2 m_{\rm e} e^2}$$ 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 5.0089 GHz H109$\alpha$ transition rate is $A_{110,109} \approx 0.3$ s$^{-1}$.

The associated natural or intrinsic line width follows from the uncertainty principle: $\Delta E \Delta t \approx \hbar$.  Substituting $h \Delta \nu$ for $\Delta E$ and $A^{-1}_{{\rm n+1}, {\rm n}}$ for $\Delta t$ for each energy level involved in the transition and summing these two uncertainties yields
$$\Delta \nu \sim A_{{\rm n+1}, {\rm n}} / \pi \sim 0.1 {\rm ~Hz}$$ This is negligibly small at the large $n$ that produce radio-frequency photons. Collisions of the emitting atoms cause collisional broadening, where $\Delta \nu$ is a small fraction of the collision rate when $\Delta n \ll n$.  Except for very large $n$, collisional broadening is usually small and the actual line profile (normalized intensity as a function of frequency) is primarily determined by Doppler shifts reflecting 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 from a source in LTE is determined by the Maxwellian speed distribution (Eq. 4B1) of atoms with mass $M$ and temperature $T$. The speed in any one coordinate of an isotropic distribution is $3^{-1/2}$ of 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
$$\vert \phi(\nu) d \nu \vert = f(v_{\rm r}) d v_{\rm r}$$ $$\phi(\nu) = \biggl( { M \over 2 \pi k T} \biggr)^{1/2} \exp \biggl[ - {M \over 2 k T} {c^2 ( \nu - \nu_0)^2 \over \nu_0^2 } \biggr] \bigg| {d v_{\rm r} \over d \nu } \bigg|$$ $$\bbox[border:3px blue solid,7pt]{\phi(\nu) = {c \over \nu_0} \biggl( { M \over 2 \pi k T} \biggr)^{1/2} \exp \biggl[ - {M c^2 \over 2 k T} {(\nu - \nu_0)^2 \over \nu_0^2 } \biggr]}\rlap{\quad \rm {(7A6)}}$$

The parameters of the normalized ($\int \phi(\nu) d \nu = 1$) line profile $\phi(\nu)$ are the center frequency $\nu_0$, the FWHM line width $\Delta \nu$, and the profile value at the center frequency $\phi(\nu_0)$.

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 ($\nu_0 = 5.0089$ GHz) in a quiescent (no macroscopic motions) HII region with temperature $T \approx 10^4$ K?

$$\Delta \nu \approx \biggl[ { 8 \ln 2\, \cdot \, 1.38 \times 10^{-16} {\rm ~erg~K}^{-1} \over ( 3 \times 10^{10} {\rm ~cm~s}^{-1})^2 } \biggr]^{1/2} \biggl( { 10^4 {\rm ~K} \over 1836 \cdot 9.11 \times 10^{-28} {\rm ~g} } \biggr)^{1/2}\cdot5.0089\times10^9{\rm ~Hz}$$ $$\Delta \nu \approx 3.6 \times 10^5 {\rm ~Hz}$$ Note that $\Delta \nu \gg A_{110,109} \approx 0.3$ Hz.

Normalization (requiring $\int \phi(\nu) d \nu = 1$) implies that the value of $\phi$ at the line center ($\nu = \nu_0$) is
$$\phi(\nu_0) = {c \over \nu_0} \biggl( { M \over 2 \pi k T} \biggr)^{1/2}$$ $$\phi(\nu_0) = {c \over \Delta \nu} \biggl( { 8 \ln 2 \, k T \over M c^2} {M \over 2 \pi k T} \biggr)^{1/2}$$ $$\bbox[border:3px blue solid,7pt]{\phi(\nu_0) = \biggl( { \ln 2 \over \pi} \biggr)^{1/2} {2 \over \Delta \nu}}\rlap{\quad \rm {(7A8)}}$$ Note that, for a given integrated (over frequency) line strength, the line strength per unit frequency at any one frequency (e.g., at $\nu_0$) is inversely proportional to the line width $\Delta \nu$.