# Frequency Redistribution in Non-Coherent Scattering

### Need to Review

<latex> \documentclass[11pt]{article} \def\inv#1Template:1 \over \def\ddtTemplate:D \over dt \def\mean#1{\left\langle #1\right\rangle} \def\sigot{\sigma_{12}} \def\sigto{\sigma_{21}} \def\eval#1{\big|_{#1}} \def\tr{\nabla} \def\dce{\vec\tr\times\vec E} \def\dcb{\vec\tr\times\vec B} \def\wz{\omega_0} \def\ef{\vec E} \def\ato{{A_{21}}} \def\bto{{B_{21}}} \def\bot{{B_{12}}} \def\bfieldTemplate:\vec B \def\apTemplate:A^\prime \def\xp{{x^{\prime}}} \def\yp{{y^{\prime}}} \def\zp{{z^{\prime}}} \def\tp{{t^{\prime}}} \def\upxTemplate:U x^\prime \def\upyTemplate:U y^\prime \def\e#1{\cdot10^{#1}} \def\hf{\frac12} \usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document}

\section*{Introduction} \hline

Figure 1. A picture of a single non-coherent scattering event. Incident light with frequency between $\nu^{\prime}$ and $\nu + d\nu^{\prime}$ and direction $\hat{n}^{\prime}$ (within an element of solid angle $d\Omega^{\prime}$), is scattered to a different frequency, between $\nu$ and $\nu + d\nu$, and direction $\hat{n}$ (into solid angle $d\Omega$), by an atom.

When a photon is scattered by an atom, it may undergo a change in frequency. Scattering events in which such {\bf frequency redistribution} occurs are said to be "{\bf non-coherent}." Hummer, Field, Spitzer, Mihalas, Avrett, and others have explored the physical causes for frequency noncoherence in the scattering of light by atoms and the effects of noncoherence arising from both the natural width of atomic levels and the thermal motion of particles in stellar atmospheres. Here we develop a mathematical formalism to generally describe the process of non-coherent scattering. We apply it to four physical cases of noncoherence.

\section*{Formulation of the Redistribution Functions} \hline The process of non-coherent {\bf line} scattering can be described very generally by a redistribution function, $\mathbb{R}$. $\mathbb{R}$ describes scattering due to a single electronic transition -- it does not describe continuum scattering or relativistic scattering. Consider the picture of non-coherent scattering presented in Figure 1. The probability $P(\nu^{\prime}, \hat{n}^{\prime};\nu, \hat{n})$ that an incident photon with frequency between $\nu^{\prime}$ and $\nu^{\prime} + d\nu^{\prime}$, traveling in direction $\hat{n}^{\prime}$ into an element of solid angle $d\Omega^{\prime}$ would be scattered to a different frequency, between $\nu + d\nu$, in direction $\hat{n}$, into $d\Omega$, is:

$$P(\nu^{\prime}, \hat{n}^{\prime};\nu, \hat{n}) = \mathbb{R}(\nu^{\prime}, \hat{n}^{\prime};\nu, \hat{n}) \;d\nu\; d\nu^{\prime}\; \frac{d\Omega}{4\pi} \; \frac{d\Omega^{\prime}}{4\pi}$$

{\bf This is the definition of a redistribution function.}

This probability is normalized to unity:

$$\int\int\int\int\!\mathbb{R}(\nu^{\prime}, \hat{n}^{\prime};\nu, \hat{n})\;d\nu \;d\nu^{\prime}\; \frac{d\Omega}{4\pi}\; \frac{d\Omega^{\prime}}{4\pi} = 1$$

For convenience, we take all integrals over frequency and /or solid angle to be complete, unless otherwise specified.

Now, integrating $\mathbb{R}$ over $\nu$ and $\Omega$ reduces $P(\nu ^{\prime}, \hat{n}^{\prime};\nu, \hat{n})$ to $P(\nu^{\prime}, \hat{n}^{\prime})$. This is the normalized probability that the atom absorbs a photon with frequency between $\nu^{\prime}$ and $\nu^{\prime} + d\nu^{\prime}$ traveling in direction $\hat{n}^\prime$ into an element of solid angle $d\Omega^\prime$, and re-emits it at {\it some} frequency in {\it some} direction. This defines the {\bf Line Profile Functions} of the transition, which could have an angular dependence. For spatially symmetric absorption profiles, this quantity is given by the familiar function, $\phi(\nu^{\prime})$:

$$\int\int\!\mathbb{R}(\nu^{\prime}, \hat{n}^{\prime};\nu, \hat{n})\; d\nu \; \frac{d\Omega}{4\pi} = \phi(\nu^{\prime})$$

By the definition of $\mathbb{R}$,

$$\int\int\!\phi(\nu^{\prime})\;d\nu^{\prime}\;\frac{d\Omega^{\prime}}{4\pi} = 1.$$

The line profile function gives a transition's normalized absorption probability as a function of $\nu^{\prime}$. We can therefore define a transition's frequency-dependent absorption coefficient, $\alpha_{\nu^{\prime}}$, in terms of the line profile function:

$$\alpha_{\nu^{\prime}} = \alpha_0\phi(\nu^{\prime})$$

where $\alpha_0$ is the line-center absorption coefficient of the transition.

\section*{Redistribution and the Radiative Transfer Equation} \hline

{\bf Figure 2.} A monochromatic radiation field with specific intensity $I_{\nu^{\prime},0}(\hat{n}^\prime)$ scattered non-coherently by a single atom. Some light is scattered to a new frequency, $\nu$, in direction $\hat{n}$. The unscattered radiation (that ultimately reaches your telescope) has specific intensity $I_{\nu^{\prime},f}(\hat{n}^\prime)$.

How does non-coherent scattering add energy to or remove energy from a beam of radiation? We now consider the effect of a single, idealized non-coherent scatterer on the spectrum of a radiation field, measured along a line of sight. Following Figure 2, define the specific intensity of the field in direction $\hat{n}^{\prime}$ before scattering to be $I_{\nu^{\prime},0}(\hat{n}^\prime)$. Let the detector lie along direction $\hat{n}^\prime$. The energy per volume per time per solid angle per frequency $(\frac{dE}{dt dV d\Omega d\nu})$ removed from the beam is given by $-\alpha_{\nu^\prime} I_{\nu^{\prime}, 0}(\hat{n^\prime})$. This follows explicitly from the definition of the absorption coefficient, $\alpha_{\nu^\prime}$.

What about light scattered non-coherently {\it into} the beam? To visualize that case, we follow Figure 3, where our detector lies along $\hat{n}$. Now, the energy per volume per time per solid angle per frequency {\bf added} into the incident beam at frequency $\nu$ is related to the {\bf amount of incident radiation at frequency $\nu^\prime$ and direction $\hat{n}^\prime$ scattered into frequency $\nu$ and direction $\hat{n}$}.

{\bf Figure 3.} A polychromatic, polydirectional radiation field with specific intensity $I_{\nu^{\prime},0}(\hat{n}^\prime)$ scattered into the line of sight of a detector at frequency $\nu$.

In the formalism of redistribution functions,

$$\frac{dE}{dt dV d\Omega d\nu} \propto \int\int \mathbb{R}(\nu^\prime, \hat{n}^\prime, \nu, \hat{n})I_{\nu^\prime}(\hat{n}^\prime)d\nu^\prime \;\frac{d\Omega^\prime}{4\pi}$$ where we have convolved the redistribution function and the specific intensity of the radiation field over frequency and solid angle to account for the continuity of the field.

This defines a "non-coherent scattering emissivity" $$j_\nu^{(s)} \propto \int\int \mathbb{R}(\nu^\prime, \hat{n}^\prime, \nu, \hat{n})I_{\nu^\prime}(\hat{n}^\prime)d\nu^\prime \;\frac{d\Omega^\prime}{4\pi}.$$

The constant of proportionality is related to the Einstein $A$'s (relative likelihoods) of other atomic transitions that may occur when an atom is excited to the level that produces the transition corresponding to $\mathbb{R}$. For example, in the case of Ly-$\beta$ (3 -> 1) scattering, an H atom in the ground state that is excited into the $n =3$ level may either undergo Ly-$\beta$ decay or H-$\alpha$ decay. The relative likelihood of H-$\alpha$ is given by

$$\epsilon = \frac{A_{H\alpha}}{A_{Ly-\beta} + A_{H\alpha}} = 0.12$$

Therefore the relative likelihood that a transition from the $n=3$ state is Ly-$\beta$ is given by $1 - \epsilon = .88$. Therefore, the non-coherent scattering emissivity is given by

$$j_\nu^{(s)} = (1-\epsilon) \int\int \mathbb{R}(\nu^\prime, \hat{n}^\prime, \nu, \hat{n})I_{\nu^\prime}(\hat{n}^\prime)d\nu^\prime \;\frac{d\Omega^\prime}{4\pi}.$$

The total emissivity of the medium is therefore the sum of the scattering emissivity and the purely "emissive" (non-scattering) emissivity $(j_\nu^{(e)})$. So the entire Radiative Transfer Equation for a medium undergoing non-coherent scattering is given by

$$\frac{dI_\nu}{ds} = -I_\nu\sum_i\alpha_{\nu,i} + j_\nu^{(e)} + \sum_i\alpha_{0,i}(1-\epsilon)\int\int\mathbb{R}(\nu^\prime, \hat{n}^\prime, \nu, \hat{n})I_{\nu^\prime}(\hat{n}^\prime)d\nu^\prime \;\frac{d\Omega^\prime}{4\pi}.$$

where we have summed over all possible upper level destinations, $i$.

\section*{Functional Forms of The Redistribution Functions I: Two Simple Cases} We now seek specific functional forms of $\mathbb{R}$. {\bf Often, but not always, we can decompose $\mathbb{R}$ into the product of an angular function $g(\hat{n}^\prime, \hat{n})$, and a frequency dependent function $F(\nu^\prime, \nu)$.} First, we consider two very simple cases that permit this simplification: {\bf coherent scattering $(\nu^\prime = \nu)$}, and {\bf complete redistribution ($(\nu^\prime, \nu)$ completely uncorrelated)}:

$$\mathbb{R}(\nu^\prime, \hat{n}^\prime, \nu, \hat{n})= g(\hat{n}^\prime, \hat{n})F(\nu^\prime, \nu)$$

Further, we consider two subcases, corresponding to two distinct functional forms of $g(\hat{n}^\prime, \hat{n})$. \begin{itemize}

• {\bf Isotropic Scattering} (e.g., the scattering of light by slow-moving dust.) In this case, $$g(\hat{n}^\prime, \hat{n}) = 1,$$ since incident radiation is equally likely to be reemitted in any direction.
• {\bf Dipole Scattering} (e.g., Thomson Scattering.) In this case, $$g(\hat{n}^\prime, \hat{n}) = \frac{3}{4}(1 + \cos^2(\theta)).$$ since light is more likely to be scattered into the dipole field.

\end{itemize} \subsection*{Case I: Coherent Scattering} We consider the case of coherent scattering, where the frequency of incident radiation is equal to the frequency of scattered radiation $(\nu^\prime = \nu)$. In this case, $F(\nu^\prime, \nu)$ is given by the Dirac Delta Function, centered on $\nu^\prime$. This is because there is a 100\% probability that incident radiation is reemitted at its original frequency. We assume $g(\hat{n}^\prime, \hat{n})$ corresponds to either of the isotropic or dipole cases. Therefore, we can write down:

$$\mathbb{R}(\nu^\prime, \hat{n}^\prime, \nu, \hat{n})= g(\hat{n}^\prime, \hat{n}) \phi(\nu^\prime) \delta(\nu^\prime -\nu).$$

The scattering emissivity is therefore given by

$$j_\nu^{(s)} = (1-\epsilon) \int\int g(\hat{n}^\prime, \hat{n}) \phi(\nu^\prime) \delta(\nu^\prime -\nu) I_{\nu^\prime}(\hat{n}^\prime) d\nu^\prime \;\frac{d\Omega^\prime}{4\pi}.$$

If the scattering is isotropic, then $g(\hat{n}^\prime, \hat{n}) = 1$, and $j_\nu^{(s)}$ becomes

$$j_\nu^{(s)} = \alpha_\nu(1-\epsilon)J_\nu.$$ \subsection*{Case II: Complete Redistribution} Complete redistribution occurs when $\nu^\prime$ and $\nu$ are completely uncorrelated. In this case, $\nu$ is determined exclusively by $\phi(\nu)$:

$$\mathbb{R}(\nu^\prime, \hat{n}^\prime, \nu, \hat{n})=g(\hat{n}^\prime, \hat{n}) \phi(\nu^\prime)\phi(\nu)$$ $$j_\nu^{(s)} = \alpha_0 (1-\epsilon) \int\int g(\hat{n}^\prime, \hat{n}) \phi(\nu^\prime)\phi(\nu) I_{\nu^\prime}(\hat{n}^\prime) d\nu^\prime \;\frac{d\Omega^\prime}{4\pi}.$$ $$= \alpha_\nu (1-\epsilon) \int\int g(\hat{n}^\prime, \hat{n}) \phi(\nu^\prime) I_{\nu^\prime}(\hat{n}^\prime) d\nu^\prime \;\frac{d\Omega^\prime}{4\pi}. $$

If the scattering is isotropic, $g(\hat{n}^\prime, \hat{n}) = 1$, so

$$ j_\nu^{(s)} = \alpha_\nu (1-\epsilon) \int \phi(\nu^\prime) J_{\nu^\prime} d\nu^\prime.$$ \subsection*{Was that realistic?} No. In reality, redistribution occurs {\bf in the frame of the atom}. Additionally, the phenomenon of {\bf Doppler redistribution} ensures that laboratory-observed redistribution is only partial. To account for these effects, we henceforth place ourselves into the rest-frame of the atom, making the following changes of variables:

Until now, we have been working in the laboratory frame. We transform to the doppler frame, where redistribution occurs in the frame of the atom.

\section*{Non-Coherent Scattering with Doppler Redistribution} \subsection*{Formalism} Let $f(\xi^\prime)$ define the absorption profile (line profile) in the atom's frame. We therefore have

$$\int \! f(\xi^\prime) \; d\xi^\prime = 1.$$

From our earlier discussion of redistribution functions, it must be the case that

$$f(\xi^\prime) \; d\xi^\prime\;\frac{d\Omega^\prime}{4\pi}$$ defines the probability that a photon with frequency between $\xi^\prime$ and $\xi^\prime+ d\xi^\prime$ is absorbed in solid angle $d\Omega^\prime$. Now, let $$p(\xi^\prime, \xi)$$ define the frequency-dependent component of the linearly-separated redistribution function, in the atom's rest-frame. Further, let $$g(\hat{n}^\prime, \hat{n})$$ define the corresponding angular function. Bringing everything together, we have that $$p(\xi^\prime, \xi)g(\hat{n}^\prime, \hat{n}) \; d\xi^\prime\;\frac{d\Omega^\prime}{4\pi}$$ gives the probability that {\it if} a photon is absorbed at $\xi^\prime, \hat{n}^\prime$, it is re-emitted at $\xi, \hat{n}$. Therefore, $$f(\xi^\prime)p(\xi^\prime, \xi)\;d\xi^\prime d\xi g(\hat{n}^\prime, \hat{n})\;\frac{d\Omega}{4\pi}\;\frac{d\Omega^\prime}{4\pi} \approx \mathbb{R}$$ for a single particle with velocity $\vec{v}$. \section*{Three Physical Cases of Non-Coherent Scattering with Doppler Redistribution} We now consider three specific cases of non-coherent scattering in the rest-frames of individual atoms. Later, we will generalize our results to entire stellar atmospheres. \subsection*{Case I: The Infinitely Sharp Atom} The first case we examine is doppler redistribution in the rest-frame of a simple two-level atom with infinitely sharp (i.e. 0 width) energy states. In this case, there is {\bf no frequency redistribution {\it in the frame of the atom}}. Instead, we have: $$f(\xi^\prime) = \delta(\xi^\prime - \nu_0)$$ $$p(\xi^\prime, \xi) = \delta(\xi^\prime - \xi)$$ \subsection*{Case II: Resonance Lines} Here we treat the case of an atom with a perfectly sharp ground state and a broadened upper state, whose finite lifetime against radiative decay leads to a Lorentzian line profile for the atom, consistent with natural broadening: $$f(\xi^\prime) = \frac{1}{\pi}\frac{\Gamma / 4\pi}{(\xi - \nu_0)^2 + (\frac{\Gamma}{4\pi})^2},$$ where $\Gamma$ is the total radiative decay rate (or "radiative damping width") of the atom, that is, $$\Gamma = \sum_{k < i} A_{ik}$$ We further assume that there are no additional perturbations of the atom while it is in its upper state. Then there will be no reshuffling of electrons among substates of the upper state, and the decay back down to the lower state will produce a photon of exactly the same frequency as the one originally absorbed. Thus we have $$p(\xi^\prime, \xi) = \delta(\xi^\prime - \xi).$$ This case applies to resonance lines in media of such low densities that collisional broadening of the upper state is completely negligible, such as the Ly-$\alpha$ line of Hydrogen in the interstellar medium. \subsection*{Case III: Complete Redistribution} The basic physical picture here is of an atom with a perfectly sharp lower state, and a broadened upper state, in a medium where collisions are so frequent that {\it all} excited electrons are randomly reshuffled over the substates of the upper state before emission occurs. The absorption profile is again the Lorentz profile, where $\Gamma$ now represents the full width (radiative plus collisional) of the upper state. In this extreme limit, the frequency of the emitted photon will have {\it no} correlation with the frequency of the absorbed photon; the probability for emission at any particular frequency is then proportional to the number of substates present at that frequency, and hence to the absorption profile itself. When complete redistribution in the atom's frame occurs we thus have$$p(\xi^\prime, \xi) = f(\xi^\prime)= \frac{1}{\pi}\frac{\Gamma / 4\pi}{(\xi - \nu_0)^2 + (\frac{\Gamma}{4\pi})^2}.$$ \section*{Solving the Three Cases: Doppler Shift Redistribution in the Laboratory Frame} In this section, we shall consider the effects of the Doppler shifts introduced by the motion of the scattering atoms relative to the laboratory frame by deriving expressions that describe the full angular and frequency dependence of redistribution in the scattering process. \subsection*{A Thermal Atmosphere} Suppose an atom moving with velocity $v < c$, which remains fixed during the scattering process, absorbs a photon ($\nu^\prime, \hat{n}^\prime$) and emits a photon ($\nu, \hat{n}$), as measured in the laboratory frame. We may therefore express the photon frequencies in the atom's frame according to: $$\xi^\prime = \nu^\prime - \nu_0(\vec{v} \cdot \hat{n}^\prime)/c$$ and $$\xi = \nu - \nu_0(\vec{v} \cdot \hat{n}) / c.$$ The redistribution function for this scattering event may be rewritten as $$\mathbb{R}_v(\nu^{\prime}, \hat{n}^{\prime};\nu, \hat{n}) = f(\underbrace{\nu^\prime - \nu_0(\vec{v} \cdot \hat{n}^\prime)/c}_{\xi^\prime})p(\underbrace{\nu^\prime - \nu_0(\vec{v} \cdot \hat{n}^\prime)/c}_{\xi^\prime}, \underbrace{\nu - \nu_0(\vec{v} \cdot \hat{n}) / c}_{\xi})g(\hat{n}^\prime, \hat{n}).$$ This is the {\it specific} redistribution function for a single species of atoms with speed $\vec{v}$. To find the {\it net} result for the entire population of atoms, we must take the average over the velocity distribution, assumed to be Maxwellian. To perform this average, we make a change of coordinates.

Diagram of the newly adopted coordinate system.

Let $\hat{n}^\prime, \hat{n}$ lie in the $\hat{n}_1, \hat{n}_2$ plane. $\hat{n}_1$ bisects $\hat{n}^\prime, \hat{n}$. Now,

$$\hat{n}^\prime = \cos(\frac{\theta}{2})\hat{n}_1 + \sin(\frac{\theta}{2})\hat{n}_2 \equiv \alpha\hat{n}_1 + \beta\hat{n}_2$$ $$\hat{n} = \cos(\frac{\theta}{2})\hat{n}_1 - \sin(\frac{\theta}{2}) \hat{n}_2 \equiv \alpha\hat{n}_1 - \beta\hat{n}_2$$

Write: $$\vec{v} = v_1\hat{n}_1 + v_2\hat{n}_2 + v_3\hat{n}_3$$ Then $$\hat{n}^\prime \cdot \vec{v} = \alpha v_1 + \beta v_2$$ $$\hat{n} \cdot \vec{v} = \alpha v_1 - \beta v_2$$ Introduce a quantity called the reduced velocity, $\vec{u}$. $$\vec{u} = \frac{\vec{v}}{v_{th}}$$ $$v_{th} = \sqrt{\frac{2kT}{m}}$$ Call the doppler width $\Delta$. $$\Delta \equiv \nu_0 \frac{v_{th}}{c} = \Delta\nu_D$$ $$\nu_0^\prime \left(\frac{\hat{n}^\prime\cdot\vec{v}}{c}\right) = \frac{\nu_0}{c}(\alpha v_1 + \beta v_2) = \Delta(\alpha u_1 + \beta u_2)$$ $$\nu_0\left(\frac{\hat{n} \cdot\vec{v}}{c}\right) = \Delta(\alpha u_1 - \beta u_2).$$ Now, the Maxwell-Boltzmann distribution can be re-expressed: $$f_{MB}(u_1, u_2, u_3)\;du_1\;du_2\;du_3 = \frac{1}{\pi^{3/2}}e^{-(u_1^2 + u_2^2 + u_3^2)}\;du_1\;du_2\;du_3$$ $$\therefore \mathbb{R}(\nu^{\prime}, \hat{n}^{\prime};\nu, \hat{n}) = \int^\infty_{-\infty}\;du_1\;\int^\infty_{-\infty}\;du_2\;\int^\infty_{-\infty}\;du_3\;f_{MB}(u_1, u_2, u_3)\cdot\mathbb{R}_u(\nu^{\prime}, \hat{n}^{\prime};\nu, \hat{n})$$ $$\mathbb{R}(\nu^{\prime}, \hat{n}^{\prime};\nu, \hat{n}) = \int^\infty_{-\infty}\;du_1\;\int^\infty_{-\infty}\;du_2\;\int^\infty_{-\infty}\;du_3\;f_{MB}(u_1, u_2, u_3)f[\nu^\prime - \Delta(\alpha u_1 + \beta u_2)]p[\nu^\prime - \Delta(\alpha u_1 + \beta u_2), \nu - \Delta(\alpha u_1 - \beta u_2)]g(\hat{n}^\prime, \hat{n}).$$ The $u_3$ integral is trivial by our choice of coordinate system. It is $\sqrt{\pi}$.

$$\therefore \mathbb{R}(\nu^{\prime}, \hat{n}^{\prime};\nu, \hat{n}) = \frac{g(\hat{n}^\prime, \hat{n})}{\pi} \times \int^\infty_{-\infty}\;du_1\;\int^\infty_{-\infty}\;du_2\;e^{-(u_1^2 + u_2^2)} f[\nu^\prime - \Delta(\alpha u_1 + \beta u_2)]p[\nu^\prime - \Delta(\alpha u_1 + \beta u_2), \nu - \Delta(\alpha u_1 - \beta u_2)]g(\hat{n}^\prime, \hat{n}).$$

\subsection*{Cases 1 and 2: Coherent Scattering in the Rest Frame of the Atom} $$p(\xi^\prime, \xi) = \delta(\xi^\prime - \xi) = \delta(\nu^\prime - \nu - 2\Delta\cdot\beta u_2)$$

To do the $u_2$ integral, let $2\Delta\beta u_2 \equiv z$. Then $$dz = 2\Delta\beta du_2$$ $$\int^{\infty}_{-\infty} e^{-u_2^2}\;du_2 f[\nu^\prime - \Delta(\alpha u_1 + \beta u_2)] \delta(\nu^\prime - \nu - 2\Delta\cdot\beta u_2)$$ $$= \frac{1}{2\Delta\beta} \int^{\infty}_{-\infty} \; dz e^{\frac{-z^2}{4\Delta^2\beta^2}}f[\nu^\prime - \Delta\alpha u_1 - \frac{z}{2}] \times \delta(\nu^\prime - \nu - z)$$ Let $z = \nu^\prime - \nu$. $$\frac{1}{2\Delta\beta}e^{\frac{-(\nu^\prime - \nu)}{4\Delta^2\beta^2}}f[\frac{1}{2}(\nu^\prime + \nu) - \Delta\alpha u_1]$$ So, finally, for coherent scattering: $$\mathbb{R}_{coh}(\nu^{\prime}, \hat{n}^{\prime};\nu, \hat{n}) = \frac{g(\hat{n}^\prime, \hat{n})}{2\pi\Delta\beta}\times e^{\frac{-(\nu^\prime - \nu)}{4\Delta^2\beta^2}}\int^\infty_{-\infty}\;du \;e^{-u^2}f[\frac{1}{2}(\nu^\prime + \nu) - \Delta\alpha u]$$

\subsubsection*{Case I: An Atmosphere of Infinitely Sharp Atoms in the Laboratory Frame} Recall that for Case I, we established that $$f(\xi^\prime) = \delta(\xi^\prime - \nu_0)$$ We therefore seek to evaluate: $$\int^\infty_{-\infty}\;du \;e^{-u^2}\delta[\frac{1}{2}(\nu^\prime + \nu) - \Delta\alpha u - \nu_0]$$ Make the following change of variables: $$y \equiv \Delta\alpha u; dy = \Delta\alpha du$$ Then the above integral becomes: $$\frac{1}{\Delta\alpha}\int^\infty_{-\infty}\;dy e^\frac{-y^2}{\Delta^2\alpha^2}\delta[\frac{1}{2}(\nu^\prime + \nu - 2\nu_0) - y]$$ $$= \frac{1}{\Delta\alpha} e^\frac{-(\nu^\prime - \nu - 2 \nu_0)^2}{4\Delta^2\alpha^2}$$ Therefore $$\mathbb{R}_I(\nu^{\prime}, \hat{n}^{\prime};\nu, \hat{n}) = \frac{g(\hat{n}^\prime, \hat{n})}{2\pi\Delta^2\alpha\beta} \times e^\frac{-(\nu^\prime - \nu)^2}{4\Delta^2\beta^2}e^\frac{-(\nu^\prime + \nu - 2\nu_0)^2}{4\Delta^2\alpha^2}$$ Therefore, although the scattering is perfectly coherent in the frame of the atom, it is non-coherent in the laboratory frame, due to the doppler effect. Now, people often rewrite this result in terms of a new set of variables, $x$ and $x^\prime$, which define a frequency shift relative to the doppler width:

$$x = \frac{1}{\Delta}(\nu - \nu_0)$$ $$x^\prime = \frac{1}{\Delta}(\nu^\prime - \nu_0)$$ $$\mathbb{R}_I(x', \hat{n}^\prime; x, \hat{n}) = \mathbb{R}_I(\nu^{\prime}, \hat{n}^{\prime};\nu, \hat{n})\underbrace{\frac{d\nu^\prime}{dx^\prime}\frac{d\nu}{dx}}_{\Delta^2}$$ $$\mathbb{R}_I(x', \hat{n}^\prime; x, \hat{n}) = \frac{g(\hat{n}^\prime, \hat{n})}{\pi\sin(\theta)}e^{-x^2-(x^\prime-x\cos(\theta))^2\csc^2(\theta)}$$ \subsubsection{Case II: An Atmosphere of Resonance Line Atoms in the Laboratory Frame} Here we have $$f(\xi^\prime) = \frac{1}{\pi}\frac{\Gamma / 4\pi}{(\xi - \nu_0)^2 + (\frac{\Gamma}{4\pi})^2}$$ Therefore, the integral over the thermal velocities becomes: $$\int^\infty_{-\infty}e^{-u^2}\frac{1}{\pi}\frac{\Gamma/4\pi}{[\frac{1}{2}(\nu^\prime + \nu - 2\nu_0) - \Delta\alpha u]^2 + (\frac{\Gamma}{4\pi})^2}$$ This integrand is similar to the Voigt function, $H(a, \nu)$. Therefore, $$R_{II}(x', \hat{n}^\prime; x, \hat{n}) = \frac{g(\hat{n}^\prime, \hat{n})}{\pi\sin(\theta)}e^{-\frac{1}{2}(x-x^\prime)^2\csc^2(\frac{\theta}{2})}H\left(a\sec(\theta/2), \frac{1}{2}(x+x^\prime)\sin(\theta/2)\right)$$ where $$a \equiv \frac{\Gamma}{4\pi\Delta}$$ and $$H(a, \nu) = \frac{a}{\pi}\int^\infty_{-\infty}\frac{e^{-y^2}dy}{(\nu - y)^2 + a^2}$$ We leave case III to the concerned reader! \end{document} <\latex>