The rotation rates and resulting line
frequencies are determined by
the quantization of angular momentum. Recall our analysis of the Bohr
atom, in which the quantization rule for the permitted electron orbital
radii
$$a_{\rm n} = { n \hbar \over m_{\rm e} v}$$
indicated that the
orbital angular momentum $L = m_{\rm e} a_{\rm n} v$ was quantized in
multiples of $\hbar \equiv h / (2 \pi)$:
$$\bbox[border:3px blue solid,7pt]{L = n \hbar}\rlap{\quad \rm
{(7D1)}}$$
The rule
that angular momentum is an integer multiple of $\hbar$ is universal
and applies to the angular momentum of a rotating molecule as well.
Consider a diatomic
molecule
containing two atoms with masses $m_{\rm
A}$ and $m_{\rm B}$ whose centers are separated by the equilibrium
distance $r_{\rm e}$. The individual atomic distances $r_{\rm A}$ and
$r_{\rm B}$
from the center of mass must obey
$$r_{\rm e} = r_{\rm A} + r_{\rm B}
\qquad {\rm and} \qquad r_{\rm A} m_{\rm A} = r_{\rm B} m_{\rm B}$$
In the center-of-mass frame,
$$L = I \omega~,$$ where $I$ is the
moment of inertia and $\omega$
is the angular velocity of the rotation.
$$L = (m_{\rm A} r_{\rm A}^2 + m_{\rm B} r_{\rm B}^2) \omega$$
$$L =
\biggl( {m_{\rm A} m_{\rm B} \over m_{\rm A} + m_{\rm B}} \biggr)
r_{\rm e}^2 \omega~.$$
$$\bbox[border:3px blue solid,7pt]{L = m
r_{\rm e}^2 \omega}\rlap{\quad \rm {(7D2)}}$$
where
$$\bbox[border:3px blue solid,7pt]{m \equiv \biggl(
{m_{\rm A} m_{\rm B} \over m_{\rm A} + m_{\rm B}} \biggr)}\rlap{\quad
\rm {(7D3)}} $$
is the reduced
mass of the molecule.
The rotational kinetic energy implied
by this
angular momentum is
$$E_{\rm rot} = {I \omega^2 \over 2} = {L^2 \over 2
I}$$
The quantization of angular momentum to integer multiples of
$\hbar$ implies that rotational energy also quantized. The
corresponding energy eigenvalues of the Schroedinger equation are:
$$\bbox[border:3px blue solid,7pt]{E_{\rm rot} = { J (J + 1) \hbar^2
\over 2 I}~, \qquad J = {\rm
0,~1,~2,} \dots }\rlap{\quad \rm {(7D4)}} $$
Note the inverse relation between
permitted rotational energies and
the moment of inertia $I$. If the upper-level rotational energy is much
higher than $kT$, few molecules will be collisionally excited to that
level and the line emission from molecules in that level will be very
weak. For example, the minimum rotational energy of the small and light
H$_2$ molecule is equivalent to a temperature $T = E_{\rm min} / k
\approx 500$ K, which is much higher than the actual temperature of
most interstellar H$_2$. Only relatively massive molecules are likely
to be detectable radio emitters in very cold (tens of K) molecular
clouds.
Quantization of rotational energy
implies that changes in
rotational energy are also quantized. The energy change of permitted
transitions is
$$\bbox[border:3px blue solid,7pt]{\Delta J = \pm 1}\rlap{\quad \rm
{(7D5)}}$$
Going from $J$ to $J -1$ releases
energy
$$ \Delta E_{\rm rot} = [ J (J + 1) - (J - 1)J] { {\hbar}^2
\over 2 I} = {\hbar^2 J \over I}~. $$
The frequency of the photon
emitted during this rotational transition is
$$\nu = {\Delta E_{\rm
rot} \over h} = {\hbar J \over 2 \pi I}~, \qquad J = {\rm 1,~2,}
...~,$$
where $J$ is the angular-momentum quantum number corresponding
to the initial energy level. In terms of the molecular constants $m$
and $r_{\rm e}$,
$$\bbox[border:3px blue solid,7pt]{\nu = { h J \over 4 \pi^2 m r_{\rm
e}^2 }~, \qquad J = {\rm 1,~2,} \dots }\rlap{\quad \rm {(7D6)}}$$
Thus a plot of the radio spectrum of
a
particular molecular species in an
interstellar cloud will look like a
ladder whose steps are all
harmonics of the fundamental frequency determined solely by the moment
of inertia of that species. The relative intensities of lines in the
ladder depend on the temperature of the cloud. Since $\nu \propto
m^{-1} r_{\rm e}^{-2}$, the lowest frequency of line emission
depends on the mass and size of the molecule. Large, heavy
molecules in cold clouds may be seen at centimeter wavelengths, but
smaller and lighter molecules emit only at millimeter wavelengths.
Example: The laboratory spectrum of
the $^{12}$C$^{16}$O
carbon-monoxide molecule shows that the fundamental $J = 1 \rightarrow
0$ transition emits a photon at $\nu = 115.271203$ GHz. (See Table 14.2
in Rohlfs & Wilson.) The distance $r_{\rm e}$ between the
C and O nuclei can be estimated from
$$r_{\rm e} = {1 \over 2 \pi}
\biggl( { h J \over m \nu} \biggr)^{1/2}$$ where the reduced mass is
$$m = {m_{\rm C} m_{\rm O} \over m_{\rm C} + m_{\rm O}} \approx m_{\rm
H} \biggl( {12 \cdot 16 \over 12 + 16} \biggr) \approx 1.67 \times
10^{-24} {\rm ~g~} \times 6.86 \approx 1.15 \times 10^{-23} {\rm
~g}~.$$
$$ r_{\rm e} = {1 \over 2 \pi} \biggl( { 6.63 \times 10^{-27}
{\rm ~erg~s} \times 1 \over 115.271203 \times 10^9 {\rm ~Hz~} \times
1.15 \times 10^{-23} {\rm ~g}} \biggr)^{1/2} \approx 1.13 \times
10^{-8} {\rm ~cm}$$
The centrifugal force increases with $J$, so the bond will stretch and $r_{\rm e}$ will increase slightly. Spectral lines emitted by more rapidly rotating $^{12}$C$^{16}$O molecules will have frequencies slightly lower than the harmonics $2 \nu_{1-0}$, $3 \nu_{1-0}$, $...$ of the $J = 1 \rightarrow 0$ line: $2 \nu_{1-0} = 230.542406$ GHz and the actual $J = 2 \rightarrow 1$ frequency is $\nu_{2-1} = 230.538001$ GHz. Chemists use these line frequencies to determine $r_{\rm e}$, and the difference between $2 \nu_{1-0}$ and $\nu_{2-1}$ is a measure of the "stiffness" of the carbon-oxygen chemical bond. Since the actual frequency is only slightly less than the harmonic frequency, the stiffness is quite high and it is clear that the fundamental vibrational frequency of the CO molecule will be much higher than the fundamental rotational frequency. Thus the vibrational line spectrum of CO begins at infrared wavelengths.
The same simple frequency formula can
be used to calculate
frequencies for rare isotopic molecules (e.g., $^{13}$C$^{16}$O) that
are more difficult to measure in the lab.
$${m(^{13}{\rm C}^{16}{\rm
O}) \over m(^{12}{\rm C}^{16}{\rm O})} = { 13 \cdot 16 / (13 + 16)
\over 12 \cdot 16 / (12 + 16)} \approx 1.0460$$
so we expect
$${
\nu_{1-0} (^{13}{\rm C}^{16}{\rm O}) \over \nu_{1-0} (^{12}{\rm
C}^{16}{\rm O}) } \approx \biggl[{m(^{13}{\rm C}^{16}{\rm O}) \over
m(^{12}{\rm C}^{16}{\rm O})} \biggr]^{-1}$$ $$ \nu_{1-0} (^{13}{\rm
C}^{16}{\rm O}) \approx 115.271203 / 1.0460 \approx 110.204 {\rm
~GHz}$$
The actual $^{13}$C$^{16}$O $J = 1 \rightarrow 2$ frequency is
110.201370 GHz.
This simple analysis shows that polar
diatomic molecules emit a
harmonic series of radio spectral lines at millimeter wavelengths.
Bigger and heavier linear polyatomic molecules have ladders of lines
starting at somewhat lower frequencies. Nonlinear molecules such as the
symmetric-top ammonia (NH$_3$) with two distinct rotational axes have
more complex spectra consisting of many parallel ladders.
Energy levels of ammonia (NH$_3$) in
the lowest vibrational state (Wilson, T. L. et al. 1993, A&A, 276,
L29). On the abscissa, $K$ is the quantum number corresponding to
the $z$-component of the angular momentum. Transitions between
the two spin states of the nitrogen atom cause the line splitting shown
and yield emission at frequencies near 24 GHz. NH$_3$ is a very useful
thermometer for molecular clouds (Ho, P. T. P., & Townes, C. H.
1983, ARA&A, 21, 239).
Molecular Excitation
Molecules are excited into $E_{\rm
rot} > 0$ states by ambient radiation and by collisions in a gas.
The minimum gas temperature $T_{\rm min}$ needed for significant
collisional excitation is roughly
$$T_{\rm min} \sim { E_{\rm rot} \over k}$$
Recall that
$$E_{\rm rot} = {J (J + 1) \hbar^2 \over 2 I} \qquad {\rm and} \qquad
\nu = {h J \over 4 \pi^2 I}$$
so
$$T_{\rm min} \sim { J (J+1) h^2 \over 2 \cdot 4 \pi^2 I k} = { h J
\over 4 \pi^2 I} { h (J + 1) \over 2 k}$$
Thus a minimum gas kinetic temperature
$$\bbox[border:3px blue solid,7pt]{T_{\rm min} \approx { \nu h (J + 1)
\over 2 k}}\rlap{\quad \rm {(7D7)}}$$
is required to excite the $J \rightarrow J -1$ transition at frequency
$\nu
$.
Example: What is the minimum gas temperature needed for significant excitation of the $^{12}$C$^{16}$O $J = 2-1$ line at $\nu \approx 230.5$ GHz?
$$T_{\rm min} \approx { \nu h (J + 1)
\over 2 k} \approx { 230.5 \times 10^9 {\rm ~Hz~} \cdot 6.63 \times
10^{-27} {\rm ~erg~s~} (2 + 1) \over 2 \times 1.3
8 \times 10^{-16} {\rm ~erg~K}^{-1} } \approx 16.6 {\rm ~K}$$
Corresponding $T_{\rm min}$ values for
other molecular transitions are listed as $E_{\rm u} / {\rm K}$ in
Table 14.2 of Rohlfs & Wilson. If $T_{\rm min} \gg 2.7$ K,
then radiative excitation
by
the cosmic microwave background is ineffective.
Line Strengths
The electric
dipole moment $\vec{d}$ of any charge distribution
$\rho(\vec{x})$ is defined as the integral
$$\vec{d} \equiv \int \vec{x} \rho(v)
dv~,$$
over the volume $v$ containing the charges. In the case of two point
charges $+q$ and $-q$ separated by $r
_{\rm e}$ on the $x$-axis,
$$ \vert d \vert = q r_{\rm e}~.$$
For molecules, $q < \vert e
\vert$. For a molecule rotating with angular velocity $\omega$, the
projection of the dipole moment perpendicular to the line of sight
varies with time as
$q r_{\rm e} \exp( - i \omega t)$. Recall that
$$ r_{\rm A} m_{\rm A
} = r_{\rm B} m_{\rm B} $$
so
$$ \dot{v}_{\rm A}= \ddot{r}_{\rm A} = \omega^2 r
_{\rm A} \qquad {\rm and} \qquad \dot{v}_{\rm B} = \ddot{r}_{\rm B} =
\omega^2 r
_{\rm B}$$
From our derivation of the Larmor formula, we found for each charge
$$E_\bot = { q \dot{v} \sin \theta \over r c^2}~,$$
where $r$ is the distance from the source. We add the contributions of
both charges to get the total
$E_\bot$:
$$ E_\bot = { q ( \omega^2 r_{\rm A} + \omega^2 r_{\rm B} ) \sin \theta
\over r c^2} \exp ( - i \omega t)$$
Thus the instantaneous power emitted is
$$P = {2 q^2 \over 3 c^3} \omega^4 \vert r_{\rm e} \exp( -i \omega t)
\vert^2~,$$
where $r_{\rm e} = r_{\rm A} + r_{\rm B}$. The time-averaged power is
$$ \langle
P \rangle = {2 q^2 \over 3 c^3} (2 \pi \nu)^4 {r_{\rm e}^2 \over 2}~.$$
$$\langle P \rangle = { 64 \pi^4 \over 3 c^3} \nu^4 \biggl( { q r_{\rm
e} \over 2} \biggr)^2$$
$$\langle P \rangle = { 64 \pi^4 \over 3 c^3} \nu^4 \vert \mu
\vert^2$$
where
$$\vert \mu \vert^2 \equiv \biggl( {q r_{\rm e} \over 2} \biggr)^2$$
defines the mean
electric dipole moment $\mu$. For a radiative
transition characterized by upper and lower levels U and L, the
spontaneous emission coefficient is
$$A_{\rm UL} = { \vert P \vert \over h \nu_{\rm UL}} $$
$$\bbox[border:3px blue solid,7pt]{A_{\rm UL} = {64 \pi^4 \over 3 h
c^3} \nu_{\rm UL}^3 \vert \mu_{\rm
UL} \vert^2}\rlap{\quad \rm {(7D8)}}$$
Notice that the spontaneous emission coefficient is proportional to
frequency cubed, so spectral lines are more prominent at the highest
radio frequencies.
Many important interstellar molecules
(e.g., H$_2$, O$_2$, etc.) are symmetric and hence have no permanent
electric dipole moments and aren't radio line sources. Those with
permanent dipole moments $\mu \sim 10^{-19}$ statcoul cm $= 0.1
$ Debye (1 Debye $\equiv 10^{-18}$ statcoul cm) have
$$\bbox[border:3px blue solid,7pt]{\vert \mu_{{\rm J+1}\rightarrow{\rm
J}} \vert^2 = {\mu^2 (J + 1)
\over (2 J + 3) }}\rlap{\quad \rm {(7D9)}}$$
(This equation reflects the complex angular wavefunctions involved, and
we won't derive it here.)
Example: Estimate the spontaneous
emission coefficient $A_{10}$ for the CO $
J = 1 \rightarrow 0$ line, given the CO dipole moment $\mu \approx
0.112$ Debye.
$$\vert \mu_{1\rightarrow0} \vert^2
\approx {(0.112 \times 10^{-18} {\rm ~statcoul~cm})^2 (0 + 1) \over (0
+ 3)} \approx 4.18 \times 10^{-39} {\rm ~statcoul
}^2 {\rm ~cm}^2$$
$$A_{10} = {64 \pi^4 \over 3 h c^3} \nu_{1-0}^3 \vert \mu_{1-0
} \vert^2$$
$$A_{10} \approx { 64 \pi^4 \cdot (115 \times 10^9 {\rm ~Hz})^2 \cdot
4.18 \times 10^{-39} {\rm ~statcoul}^2 {\rm ~cm}^2 \over 3 \cdot 6.63
\times 10^{-27} {\rm ~erg~s} \cdot (3 \times 10^{10} {\rm
~cm~s}^{-1})^2 }$$
$$A_{10} \approx 7.4 \times 10^{-8} {\rm ~s}^{-1} \sim 2.3 {\rm
~yr}^{-1}$$
The typical time $A_{10}^{-1} \approx 10^7$ s to emit one photon
spontaneously is much longer than the average time
between molecular collisions in an interstellar molecular cloud with
density $10^2$ to $10^3$ H$_2$ molecules cm$^{-3}$, so the
CO can approach LTE with the excitation temperature of the $J=
1\rightarrow0$ line being nearly equal to the kinetic temperature of
the molecular cloud. For any molecular transition, there is a
critical density
$n^*$ at
which the collisional rate $n$(H$_2$)$\sigma v$ equals $A$.
Typically
$\sigma \sim 10^{-16}$ cm$^2$ and $v \sim 10^5$ cm
s$^{-1}$. Transitions with high emission coefficients (e.g., the
HCN line at 88.63 GHz has $A \approx 2.0 \times 10^{-5}$ s$^{-1}$) are
collisionally excited only at very high densities.
Formation of
interstellar glycoaldehyde, a sugar consisting of two
carbon atomes, two oxygen atoms, and four hydrogen atoms. This
pre-biotic molecule is a building block for DNA and RNA.
Image
credit
Red-
and blue-shifted CO outflows from
a young stellar object in the dark cloud L1157.
Image
credit
The
protostar NGC 1333 IRS 4A (gray
spot) has ejected two supersonic jets whose Doppler velocities are
indicated by the false colors (blue = approaching, red = receding) in
this VLA image of a 43 GHz SiO (silicon monoxide) line. SiO is an
excellent tracer of shocked molecular gas.
Image
credit
CO $J = 1 \rightarrow 0$ emission from
a shell ejected by the evolved carbon star TT Cyg, imaged by the
Plateau de Bure interferometer (Olofsson et al. 2000, A&A, 353,
583).
The CO $J = 1 \rightarrow 0$ contours
superimposed on a gray-scale optical image of NGC 5194 trace the spiral
arms and outline regions in which stars are forming (Regan, M. W. et
al. 2001, ApJ, 561 218).
The circumnuclear water maser in NGC
4258.
Image
credit
Keplerian orbits of the circumnuclear
water masers in NGC 4258.
Image
credit
This
VLA image of the 346 GHz CO $J=3\rightarrow 2$ emission from the z =
6.42 quasar J1148+5251
redshifted to 46.61 GHz (Walter et al. 2004, ApJ, 615, L17) reflects
the evolution of stars and supermassive black holes when the universe
was $< 10^9$ years old.