Line Radiative Transfer

We used Larmor's equation to estimate the spontaneous emission coefficients $A_{\rm UL}$ for recombination lines. $A_{\rm UL}$ is just the photon emission rate (s$^{-1}$) for an undisturbed atom or molecule going from an upper (U) to lower (L) energy state. $B_{\rm UL}$ is the stimulated emission coefficient, and $B_{\rm LU}$ is the absorption coefficient.  Einstein showed that the absorption and stimulated-emission coefficients can be calculated from the spontaneous emission coefficient.

If this system is in its lower energy state, it may absorb a photon of frequency $\nu_0$ and go to the upper state. The rate for this process is proportional to the mean spectral energy density $\bar{U}_\nu$ (e.g., erg cm$^{-3}$ Hz$^{-1}$) of the radiation field, so the Einstein absorption coefficient $B_{\rm LU}$ is defined such that the product
$$\bbox[border:3px blue solid,7pt]{B_{\rm LU} \bar{U}_\nu}\rlap{\quad \rm {(7B1)}}$$
is the rate (s$^{-1}$) of photon absorption by a single atomic or molecular system in its lower energy state.

Einstein's important insight was that there must be a third process in addition to spontaneous emission and absorption. It is stimulated emission, in which a photon of energy $h \nu_0$ stimulates the system in the upper energy state to emit a second photon with the same energy $h \nu_0$. The rate of this process is also proportional to $\bar{U}_\nu$, so by analogy we define the Einstein stimulated-emission coefficient $B_{\rm UL}$ such that the product
$$\bbox[border:3px blue solid,7pt]{B_{\rm UL} \bar{U}_\nu}\rlap{\quad \rm {(7B2)}}$$
is the rate (s$^{-1}$) of stimulated photon emission by a single quantum system in its upper energy state. Stimulated emission is sometimes called negative absorption. It is not intuitively familiar because it is weak in cool objects at visible wavelengths, but it is very important at radio wavelengths where $h \nu / ( k T) \ll 1$.

Suppose we have a macroscopic physical system containing a large number of atoms or molecules in full thermodynamic equilibrium (TE).  TE is a stationary state, so the average rates of photon creation and destruction are equal. If the macroscopic system contains $(N_{\rm U},\,N_{\rm L})$ atoms or molecules in the (upper, lower) energy states per unit volume, then the balance of photon creation by spontaneous emission or stimulated emission and photon destruction by absorption implies
$$\bbox[border:3px blue solid,7pt]{N_{\rm U} A_{\rm UL} + N_{\rm U} B_{\rm UL} \bar{U}_\nu = N_{\rm L} B_{\rm LU} \bar{U}_\nu}\rlap{\quad \rm {(7B3)}}$$

In TE, the ratio of $N_{\rm U}$ to $N_{\rm L}$ is fixed by the Boltzmann equation:
$${ N_{\rm U} \over N_{\rm L}} = {g_{\rm U} \over g_{\rm L}} \, \exp \biggl[ - {(E_{\rm U} - E_{\rm L}) \over k T} \biggr] = {g_{\rm U} \over g_{\rm L}} \, \exp \biggl( - {h \nu_0 \over k T} \biggr)~,$$
where $g_{\rm U}$ and $g_{\rm L}$ are the numbers of states with energies $E_{\rm U}$ and $E_{\rm L}$. The quantities $g_{\rm U}$ and $g_{\rm L}$ are called the statistical weights of those energy states. Examples of statistical weights include:

(1) Hydrogen atoms have $g_{\rm n} = 2 n^2$, where ${n} = 1,~2,~3,\, ...$ is the electronic energy level. The number $2n^2$ is the product of the 2 electron spin states and $n^2$ orbital angular momentum states in the $n$th energy level.

(2) Rotating linear molecules (e.g., carbon monoxide, CO) have $g = 2J +1$, where $J = 0,~1,~2,\, ...$ is the angular momentum quantum number. For each $J$, there are $2J + 1$ possible values of the $z$-component of the angular momentum: $J_z = -J,~-(J - 1),~...,~-1,~0,~1,~...,~(J-1),~J$.

(3) Hydrogen atoms have two hyperfine energy levels whose difference yields the $\lambda = 21$ cm ($\nu_0 = 1420.406\dots$ MHz) HI line; $g_{\rm U} = 3$ and $g_{\rm L} = 1$.

From the balance of photon creation and destruction we can extract the photon spectral energy density
$$\bar{U}_\nu = { N_{\rm U} A_{\rm UL} \over N_{\rm L} B_{\rm LU} - N_{\rm U} B_{\rm UL} } = {A_{\rm UL} \over (N_{\rm L} / N_{\rm U})B_{\rm LU} - B_{\rm UL} }$$

Full TE implies both
$$ \bar{U}_\nu = A_{\rm UL} \biggl[ {g_{\rm L} \over g_{\rm U}} \exp\biggl( {h \nu_0 \over k T} \biggr) B_{\rm LU} - B_{\rm UL} \biggr]^{-1}$$
and
$$\bar{U}_\nu = {4 \pi B_\nu(\nu_0, T) \over c}$$
Inserting the Planck radiation law for $B_\nu(\nu_0, T)$ gives $$\bar{U}_\nu = {4 \pi \over c} { 2 h \nu_0^3 \over c^2} \biggl[ \exp \biggl( { h \nu_0 \over k T} \biggr) - 1 \biggr]^{-1}~.$$
Next we equate these two expressions for $\bar{U}_\nu$ at the line center frequency:
$$A_{\rm UL} \biggl[ {g_{\rm L} \over g_{\rm U}} \exp\biggl( {h \nu_0 \over k T} \biggr) B_{\rm LU} - B_{\rm UL} \biggr]^{-1} = {4 \pi \over c} { 2 h \nu_0^3 \over c^2} \biggl[ \exp \biggl( { h \nu_0 \over k T} \biggr) - 1 \biggr]^{-1}~.$$
This equality holds for all temperatures $T$, so
$$ {A_{\rm UL} \over B_{\rm UL}} \biggl[ {g_{\rm L} \over g_{\rm U}} {B_{\rm LU} \over B_{\rm UL}} \exp\biggl({ h \nu_0 \over k T}\biggr) - 1 \biggr]^{-1} = {8 \pi h \nu_0^3 \over c^3} \biggl[ \exp \biggl( { h \nu_0 \over k T} \biggr) - 1 \biggr]^{-1}$$
implies both
$$\bbox[border:3px blue solid,7pt]{{ g_{\rm L} \over g_{\rm U} } {B_{\rm LU} \over B_{\rm UL} } = 1}\rlap{\quad \rm {(7B4)}}$$
and
$$\bbox[border:3px blue solid,7pt]{{A_{\rm UL} \over B_{\rm UL}} = {8 \pi h \nu_0^3 \over c^3}}\rlap{\quad \rm {(7B5)}}$$

These two equations relate the three quantities $A_{\rm UL}$, $B_{\rm LU}$, and $B_{\rm UL}$, so all three can be computed if only one (e.g., the spontaneous emission coefficient $A_{\rm UL}$) is known.  They also prove that $B_{\rm UL}$ is not zero; that is, spontaneous emission must occur.

These equations are valid for any microscopic physical system. They just relate constants characteristic of individual atoms or molecules for which the macroscopic statistical concepts of TE or LTE are meaningless. Even though TE was used to motivate the derivation, the dependences on temperature $T$ and frequency $\nu$ dropped out for a line at a single frequency $\nu_0$. Thus these equations are valid for systems in TE or LTE but are not dependent on TE or LTE. [Recall our derivation of Kirchoff's law, which also made use of full TE but which yielded
$$ {\epsilon_\nu (T) \over \kappa_\nu (T)} = B_\nu(T)$$
relating the emission and absorption coefficients of any matter in LTE, independent of the actual radiation field.]

Quantum Radiative Transfer

We can use the two equations (7B4 and 7B5) relating the three Einstein coefficients to solve the spectral-line radiative transfer problem in terms of the spontaneous emission coefficient $A_{\rm UL}$ alone. Recall the radiative transfer equation:
$$ {d I_\nu \over ds} = -\kappa_\nu I_\nu + \epsilon_\nu$$
where $I_\nu$ is the specific intensity and has dimensions
$${\rm dim}(I_\nu) = {{\rm energy/time} \over {\rm frequency} \cdot {\rm solid~angle} \cdot {\rm area}}~.$$

Also $${\rm area} = d \sigma = {d v \over d s}~,$$
where $d \sigma$ is the cross sectional area and $d v$ is the volume element.
$${d I_\nu \over ds} = {d E(\nu) / dt \over d \nu d \Omega d v}$$
$${d I_\nu \over ds} = {d E(\nu) \over d \nu d t d \Omega d v}$$
The total change in energy $d E(\nu)$ at frequency $\nu$ is the sum of the energy changes from (1) spontaneous emission, (2) absorption, and (3) stimulated emission:
$$d E(\nu) = d E_{\rm e}(\nu) + d E_{\rm a}(\nu) + d E_{\rm s}(\nu)$$ These three energy terms are
$$d E_{\rm e}(\nu) = (h \nu_0) (N_{\rm U} d v) (A_{\rm UL} dt) [\phi(\nu) d \nu] {d \Omega \over 4 \pi}$$
$$d E_{\rm a}(\nu) = - (h \nu_0) (N_{\rm L} d v) (B_{\rm LU} d t) \bar{U}_\nu [\phi(\nu) d \nu] {d \Omega \over 4 \pi}$$
$$d E_{\rm s}(\nu) = (h \nu_0) (N_{\rm U} d v) (B_{\rm UL} d t) \bar{U}_\nu [\phi(\nu) d \nu] {d \Omega \over 4 \pi}~,$$
where $\phi(\nu)$ is the normalized line profile. Thus
$$ { d I_\nu \over ds} = -{h \nu_0 \over c} (N_{\rm L} B_{\rm LU} - N_{\rm U} B_{\rm UL}) I_\nu \phi(\nu) + {h \nu_0 \over 4 \pi} N_{\rm U} A_{\rm UL} \phi(\nu) = -\kappa_\nu I_\nu + \epsilon_\nu$$
$$ \kappa_\nu = {h \nu_0 \over c} (N_{\rm L} B_{\rm LU} - N_{\rm U} B_{\rm UL} ) \phi(\nu)$$

The final results are
$$\kappa_\nu = {h \nu_0 \over c} N_{\rm L} B_{\rm LU} \biggl( 1 - {N_{\rm U} \over N_{\rm L}}{g_{\rm L} \over g_{\rm U}} \biggr) \phi(\nu)$$
and $$\epsilon_\nu = { h \nu_0 \over 4 \pi} N_{\rm U} A_{\rm UL} \phi(\nu)~.$$

The ratio of these emission and absorption coefficients is
$${\epsilon_\nu \over \kappa_\nu} = { N_{\rm U} A_{\rm UL} c \over 4 \pi N_{\rm L} B_{\rm LU}} \biggl( 1 - {N_{\rm U} \over N_{\rm L}}{g_{\rm L} \over g_{\rm U}} \biggr)^{-1}$$
$${\epsilon_\nu \over \kappa_\nu} = {N_{\rm U} (8 \pi h \nu_0^3 / c^2) B_{\rm UL} \over 4 \pi N_{\rm L} B_{\rm LU}} \biggl( 1 - {N_{\rm U} \over N_{\rm L}}{g_{\rm L} \over g_{\rm U}} \biggr)^{-1}$$ $${\epsilon_\nu \over \kappa_\nu} = {2 h \nu_0^3 \over c^2} {B_{\rm UL} \over B_{\rm LU}} \biggl( {N_{\rm L} \over N_{\rm U}} - {g_{\rm L} \over g_{\rm U}} \biggr)^{-1}$$
$${\epsilon_\nu \over \kappa_\nu} = {2 h \nu_0^3 \over c^2} \biggl( {g_{\rm U} \over g_{\rm L}} {N_{\rm L} \over N_{\rm U}} - 1 \biggr)^{-1}$$

In LTE, Kirchoff's law independently implies
$${\epsilon_\nu \over \kappa_\nu} = B_\nu(T) = {2 h \nu_0^3 \over c^2} \biggl[ \exp \biggl( {h \nu \over k T} \biggr) - 1 \biggr]^{-1}$$
so
$${g_{\rm U} \over g_{\rm L}} {N_{\rm L} \over N_{\rm U}} = \exp \biggl( { h \nu_0 \over k T} \biggr)$$
and we recover the Boltzmann distribution for LTE (not just for full TE):
$$\bbox[border:3px blue solid,7pt]{{N_{\rm U} \over N_{\rm L}} = {g_{\rm U} \over g_{\rm L}} \exp \biggl( - { h \nu_0 \over k T} \biggr)}\rlap{\quad \rm {(7B6)}}$$

Using
$$\kappa_\nu = {h \nu_0 \over c} N_{\rm L} B_{\rm LU} \biggl( 1 - {N_{\rm U} \over N_{\rm L}} {g_{\rm L} \over g_{\rm U}} \biggr) \phi(\nu)$$
and the assumption of LTE, we can substitute
$$B_{\rm LU} = {g_{\rm U} \over g_{\rm L}} B_{\rm UL} = {g_{\rm U} \over g_{\rm L}} {A_{\rm UL} c^3 \over 8 \pi h \nu_0^3}$$
and
$${N_{\rm U} \over N_{\rm L}} {g_{\rm L} \over g_{\rm U}} = \exp \biggl( - {h \nu_0 \over k T} \biggr)$$
to get the line opacity coefficient
$$\bbox[border:3px blue solid,7pt]{\kappa_\nu = {c^2 \over 8 \pi \nu_0^2} {g_{\rm U} \over g_{\rm L}} N_{\rm L} A_{\rm UL} \biggl[ 1 - \exp \biggl( - {h \nu_0 \over k T} \biggr) \biggr] \phi(\nu) {\rm \qquad (in~LTE)}}\rlap{\quad \rm {(7B7)}}$$
in terms of the spontaneous emission rate $A_{\rm UL}$ only; the stimulated emission coefficient $B_{\rm UL}$ and absorption coefficient $B_{\rm LU}$ have been eliminated.

The quantity
$$ \biggl[ 1 - \exp \biggl( - {h \nu_0 \over k T} \biggr) \biggr]$$
in the line opacity equation above has two terms. The positive term (1) comes from absorption and the negative exponential term represents the negative opacity of stimulated emission. In the radio limit $h \nu_0 \ll k T$,
$$ \biggl[ 1 - \exp \biggl( - {h \nu_0 \over k T} \biggr) \biggr] \approx {h \nu_0 \over k T} \ll 1 $$
Thus stimulated emission nearly cancels absorption and reduces line opacity significantly. Also note that, in the radio limit, $\kappa_\nu \propto T^{-1}$. In the Rayleigh-Jeans limit, $\kappa_\nu B_\nu \propto T^0$; that is, the brightness of an optically thin ($\tau \ll 1$) radio emission line may be proportional to the column density of emitting gas but nearly independent of its temperature.

Even if a macroscopic system is not in LTE, we can define its excitation temperature $T_{\rm e}$ by
$$\bbox[border:3px blue solid,7pt]{{N_{\rm U} \over N_{\rm L}} \equiv {g_{\rm U} \over g_{\rm L}} \exp \biggl( - { h \nu_0 \over k T_{\rm e}} \biggr)}\rlap{\quad \rm {(7B8)}}$$
If for some reason the upper level is overpopulated; that is
$${N_{\rm U} \over N_{\rm L}} > {g_{\rm U} \over g_{\rm L}}~,$$
then $T_{\rm e}$ is negative and
$$\biggl[ 1 - \exp \biggl( - {h \nu_0 \over k T_{\rm e}} \biggr) \biggr]$$
is negative, giving a negative opacity coefficient $\kappa_\nu$ (gain instead of loss), or maser (microwave amplification by stimulated emission of radiation) amplification. Astrophysical masers are common at radio frequencies.  They can have extremely high brightness temperatures, much higher than the kinetic temperature of the masing gas.