Line Radiative Transfer

Einstein Coefficients

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. To solve the spectral-line radiative transfer problem, we also need the stimulated emission coefficient $B_{\rm UL}$ and the absorption coefficient $B_{\rm LU}$.  Einstein showed that both the stimulated-emission and absorption coefficients can be calculated from the spontaneous emission coefficient.



The Einstein coefficients for a two-level system: $A_{\rm UL}$ for spontaneous emission, $B_{\rm LU}$ for absorption, and $B_{\rm UL}$ for stimulated emission.


Consider any two energy levels $E_{\rm U}$ and $E_{\rm L}$ of a quantum system such as a single atom or molecule.  The photon emitted or absorbed during a transition between the upper and lower states will have energy
$$E = E_{\rm U} - E_{\rm L}$$ and contribute to a spectral line with frequency $\nu_0 = E / h$.  The energy levels actually have small but finite widths, so the spectral line has some narrow line profile $\phi(\nu)$ centered on $\nu = \nu_0$ and conventionally normalized such that $\int_0^\infty \phi(\nu) d \nu = 1$.

If this system is in its lower energy state, it may absorb a photon of frequency $\nu \approx \nu_0$ and go to the upper state. The rate for this process is proportional to the profile-weighted mean energy density
$$\bar{U} \equiv \int_0^\infty U_\nu \,\phi(\nu) d\nu$$ of the radiation field, so the Einstein absorption coefficient $B_{\rm LU}$ is defined to make the product
$$\bbox[border:3px blue solid,7pt]{B_{\rm LU} \bar{U}}\rlap{\quad \rm {(7B1)}}$$ equal the rate (s$^{-1}$) of photon absorption from the radiation field by a single atomic or molecular system in its lower energy state.

Einstein's critical 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 and direction. The rate for this process is also proportional to $\bar{U}$, so by analogy with Equation 7B1 the Einstein stimulated-emission coefficient $B_{\rm UL}$ is defined to make the product
$$\bbox[border:3px blue solid,7pt]{B_{\rm UL} \bar{U}}\rlap{\quad \rm {(7B2)}}$$ equal the rate (s$^{-1}$) of stimulated photon emission from a single quantum system in its upper energy state. Stimulated emission is sometimes called negative absorption. It is not intuitively familiar because negative absorption is much weaker than ordinary absorption in room-temperature objects at visible wavelengths, but it competes effectively with ordinary absorption 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) with the surrounding radiation field.  TE is a stationary state, so the average rates of photon creation and destruction must be equal. If the macroscopic system contains $(N_{\rm U},\,N_{\rm L})$ atoms or molecules per unit volume in the (upper, lower) energy states, then the balance of photon creation by spontaneous emission or stimulated emission and photon destruction by absorption implies
$$\bbox[border:3px blue solid,7pt]{N_{\rm U} A_{\rm UL} + N_{\rm U} B_{\rm UL} \bar{U} = N_{\rm L} B_{\rm LU} \bar{U}}\rlap{\quad \rm {(7B3)}}$$

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 $g_{\rm L} = 1$.

The balance of photon creation and destruction in TE connects the mean energy density of blackbody radiation to properties of the quantum system (atom or molecule):
$$\bar{U} = { N_{\rm U} A_{\rm UL} \over N_{\rm L} B_{\rm LU} - N_{\rm U} B_{\rm UL} } = {A_{\rm UL} \over (N_{\rm L} / N_{\rm U})B_{\rm LU} - B_{\rm UL} }$$ Full TE at temperature $T$ implies both
$$ \bar{U} = A_{\rm UL} \biggl[ {g_{\rm L} \over g_{\rm U}} \exp\biggl( {h \nu_0 \over k T} \biggr) B_{\rm LU} - B_{\rm UL} \biggr]^{-1}$$ and
$$\bar{U} = {4 \pi \over c}\int_0^\infty B_\nu(T) \,\phi(\nu) d \nu~.$$ Inserting the Planck radiation law for $B_\nu(T)$ near $\nu = \nu_0$ gives $$\bar{U} \approx {4 \pi \over c} { 2 h \nu_0^3 \over c^2} \biggl[ \exp \biggl( { h \nu_0 \over k T} \biggr) - 1 \biggr]^{-1}~.$$
Next we equate these two expressions for $\bar{U}$ 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 are called the equations of detailed balance.  They relate $A_{\rm UL}$, $B_{\rm LU}$, and $B_{\rm UL}$, so all three quantities can be computed if only one (e.g., the spontaneous emission coefficient $A_{\rm UL}$) is known.  Equations 7B4 and 7B5 also prove that $B_{\rm UL}$ is not zero; that is, stimulated emission must occur.

Note that these equations are valid for any microscopic physical system because they 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 also valid for macroscopic systems whether or not they are in TE or LTE. [Recall the 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. The radiative transfer equation (Eq. 2B4) is:
$$ {d I_\nu \over ds} = -\kappa_\nu I_\nu + \epsilon_\nu~,$$ where $I_\nu$ is the specific intensity, $\kappa_\nu$ is the net fraction of photons absorbed (the difference between ordinary absorption and negative absorption) per unit length, and $\epsilon_\nu$ is the volume emission coefficient.




Three processes must be considered: (1) absorption from the lower to upper level, (2) stimulated emission, which we treat as negative absorption from the upper to lower level, and (3) spontaneous emission.  They contribute the three terms in the equation below:
$$ { d I_\nu \over ds} = -{h \nu_0 \over c} [N_{\rm L} B_{\rm LU}I_\nu \,\phi(\nu)]- {h \nu_0 \over c} [-N_{\rm U} B_{\rm UL} I_\nu \,\phi(\nu)] + {h \nu_0 \over 4 \pi} N_{\rm U} A_{\rm UL} \phi(\nu) = -\kappa_\nu I_\nu + \epsilon_\nu$$ The net absorption coefficient is
$$ \kappa_\nu = {h \nu_0 \over c} (N_{\rm L} B_{\rm LU} - N_{\rm U} B_{\rm UL} ) \phi(\nu)$$ Equation 7B4 can be used to eliminate the stimulated emission coefficient $B_{\rm UL}$ to yield
$$\kappa_\nu = {h \nu_0 \over c} N_{\rm L} B_{\rm LU} \biggl( 1 - {N_{\rm U} \over N_{\rm L}}{g_{\rm L} \over g_{\rm U}} \biggr) \phi(\nu)$$ The emission coefficient is $$\epsilon_\nu = { h \nu_0 \over 4 \pi} N_{\rm U} A_{\rm UL} \phi(\nu)~.$$

The ratio of these emission and (net) absorption coefficients is
$${\epsilon_\nu \over \kappa_\nu} = {c N_{\rm U} A_{\rm UL} \over 4 \pi N_{\rm L} B_{\rm LU}} \biggl( 1 - {N_{\rm U} \over N_{\rm L}}{g_{\rm L} \over g_{\rm U}} \biggr)^{-1}$$ Equation 7B5 can be used to eliminate $A_{\rm UL}$: $${\epsilon_\nu \over \kappa_\nu} = {N_{\rm U} (8 \pi h \nu_0^3 / c^2) B_{\rm UL} \over 4 \pi N_{\rm L} B_{\rm LU}} \biggl( 1 - {N_{\rm U} \over N_{\rm L}}{g_{\rm L} \over g_{\rm U}} \biggr)^{-1}$$ $${\epsilon_\nu \over \kappa_\nu} = {2 h \nu_0^3 \over c^2} {B_{\rm UL} \over B_{\rm LU}} \biggl( {N_{\rm L} \over N_{\rm U}} - {g_{\rm L} \over g_{\rm U}} \biggr)^{-1}$$ Finally, Equation 7B4 can be used to eliminate both $B_{\rm UL}$ and $B_{\rm LU}$: $${\epsilon_\nu \over \kappa_\nu} = {2 h \nu_0^3 \over c^2} \biggl( {g_{\rm U} \over g_{\rm L}} {N_{\rm L} \over N_{\rm U}} - 1 \biggr)^{-1}$$

In LTE, Kirchoff's law independently implies
$${\epsilon_\nu \over \kappa_\nu} = B_\nu(T) = {2 h \nu^3 \over c^2} \biggl[ \exp \biggl( {h \nu \over k T} \biggr) - 1 \biggr]^{-1}$$ 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 Rayleigh-Jeans 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 significantly reduces the net line opacity. Since $\kappa_\nu \propto T^{-1}$, $\kappa_\nu B_\nu \propto T^0$.  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 $T_{\rm x}$ by
$$\bbox[border:3px blue solid,7pt]{{N_{\rm U} \over N_{\rm L}} \equiv {g_{\rm U} \over g_{\rm L}} \exp \biggl( - { h \nu_0 \over k T_{\rm x}} \biggr)}\rlap{\quad \rm {(7B8)}}$$ 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 $T_{\rm x}$ is actually negative, $$\biggl[ 1 - \exp \biggl( - {h \nu_0 \over k T_{\rm x}} \biggr) \biggr]$$ is negative, and Equation 7B7 gives a negative net opacity coefficient $\kappa_\nu$. Negative opacity implies brightness gain instead of loss.  At radio wavelengths this is called maser (microwave amplification by stimulated emission of radiation) amplification. Astrophysical masers are common at radio frequencies because $h\nu \ll kT$.  They can have brightness temperatures as high as $10^{15}$ K, much higher than the kinetic temperature of the masing gas.