Talk 3: Radiative Transfer, Application to 21cm Global Signal, and the WF Effect

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\title{Adrian Liu's Summer 2015 Global Signal Talks: Part 3: Radiative Transfer and 21cm Global Signal + Wouthuysen Field Effect with Guest Lecturer Yunfan Zhang} \author{Notes by Cherie Day} \date{6/25/2015}

\begin{document} \maketitle


\section{Principles of Radiative Transfer} \begin{itemize} \item Radiative transfer is discussed in detail on Astrobaki, here we mean to give a brief overview through notes of the talk. \item Quest: How do we quantify radiation? We want to know how much radiation is there, but what do we mean by that? \item Answer: Often, we mean how much radiation is at a given place, going in a certain direction.This is known as the intensity: \begin{equation} I(x, \Omega, \nu) = {dE \over {dAdtd \Omega}} \end{equation} where $d\Omega$ is a small chunk of a solid angle, see Figure below, and $dE$, $dA$, and $dt$ are smalls amounts of energy, area of the detector, and time, respectively. \begin{figure}[!h] \centering \includegraphics[width=\textwidth]{SolidAngle_GSnotes.png} \caption{Illustration of a solid angle ($\Omega$) through a slice of volume, $dV$. In spherical coordinates, the differential, $d\Omega$, is defined as $d\Omega = \sin\theta d\theta d\phi$, where $\theta$ is measured from the North pole (often the z-axis) and $d\phi$ is the longitude (on an xyz coordinate system, from the x to y axes). \label{solidangl}} \end{figure} \item Photons of different frequency propagate differently, thus we define the Specific Intensity (the intensity per frequency bin) as \begin{equation} I_{\nu} = {dI \over d\nu} = {dE \over {dAdtd \Omega d\nu}} \end{equation} \item Having defined the quantity of interest, we want to establish a differential equation that governs its evolution.

This is called the Equation of Radiative Transfer. To write it down, we simply note that there are processes that increase $I_\nu$ and processes that decrease it, and that the amount of decrease is proportional to the amount we start with (in other words each photon experiences the statistical average).

\begin{equation} \label{partnu_parts} {\partial I_{\nu} \over \partial s} = {-\alpha_{\nu}I_{\nu} + j_{\nu}} \end{equation}

What are the processes that change $I_{\nu}$? \\

Going along a direction $s$, $\alpha_{\nu}$ is the out-scattering and absorption and $j_{\nu}$ is the in-scattering and emission. \item Do not be fooled by the apparent simplicity of this formula. We haven't tapped into the physical processes responsible and thus have no idea so far what $j_{\nu}$ and $\alpha_\nu$ are. The entire field of Quantum Electrodynamics has been developed to compute these scattering and absorption coefficients. \item You may also see scattering into the beam (increases $I_{\nu}$ in direction $s$) written by the form: \begin{equation} j_{\nu}: {\int I_{\nu'}(\Omega')W(\nu \Omega | \nu' \Omega')d\nu' d\Omega'} \end{equation} where $W$ is called the transfer function. Then again we know little about W, except that it has to be symmetric. And even if we are given the form of W, the dependence of $j_\nu$ on $I_\nu$ often makes solving the equation analytically impossible and numerically beyond the reach of even the best of current computational resources (note this is because we are in a six dimensional space). For a discussion of the same equation applied to neutrinos, and the numerical challenges involved, see Zhang 2013 (PhysRevD.88.105009). \item Define optical depth: $d\tau \equiv \alpha_{\nu}ds$. How many times will you scatter if you shoot a photon into a box? \begin{equation} \text{number of times photon is scattered} = ds \cdot \sigma \cdot n_s \end{equation} where $ds$ is the length traveled through, $\sigma$ is the cross-section, and $n_s$ is the number density of scatterers. This is the optical depth: \begin{equation} d\tau = {n \cdot \sigma \cdot ds} = {ds \over l_{mfp}} \end{equation} where $mfp$ is the mean free path (i.e. the average distance a photon can travel before being scattered). From this we can see: \begin{equation} \tau = {\int d\tau} = {\int {ds \over l_{mfp}}} = \text{number of mfp in your path s} \end{equation} \item Dividing both sides of the derivative with respect to s equation by $\alpha_{\nu}$, we obtain \begin{equation} {dI_{\nu} \over d\tau} = {-I_{\nu} + S_{\nu}} \end{equation} where $S_{\nu}$ is called the source function and has the same units as $I_{\nu}$ and $d\tau$ is dimensionless. \item However, there are cases when the equation is simple and can be solved by elementary means. One such case is the in the condition of Local Thermodynamic Equilibrium (LTE). We know in this case that the source function is given by Planck's blackbody curve: \begin{equation} S_{\nu} = B_{\nu} = { {{2h \nu^3} \over c^2} {1 \over {e^{{h\nu} \over kT} -1}}} \end{equation} \item Even in its general form, the radiative transfer equation far from describes all properties of radiation. What are some of its limitations? Note that it is a classical treatment since it ignores quantum effects. It also ignores time (i.e. the changes in time). Also, it doesn't include polarization or interference. \item Lets try to make the left hand side a little more general. We start by allowing explicit time dependence: \begin{equation} \label{lotsofpartials} {{1 \over c}{\partial \over \partial t}{I_{\nu}+}{\partial \over \partial s}{I_\nu}} = {-\alpha_{\nu} I_{\nu} + j_{\nu}} \end{equation} Is the left hand side now complete? If $\alpha_{\nu}$ and $j_{\nu}$ are zero (i.e. in the case of free streaming photons), then the left hand side must also be zero. This, in physics, is called the ``continuity equation". Recall that $I_\nu$ also depends on frequency and direction, or in other words on the momentum of photons, and thus, a change in the momentum would render the left hand side non-zero. You may ask what processes may change the momentum of photons, the answers of which include gravitational redshift and the universe's expansion. We thus need to include a momentum term. \item More precisely, consider the groups $(dAdt)$ and $(d\Omega d\nu)$: \begin{equation} {dE \over {(dAdt)(d\Omega d \nu)}} \Longleftrightarrow { {\partial E} \over {{\partial^3 x}{\partial^3 p}}} \end{equation} up to a constant. So $I_{\nu}$ is secretly a phase space distribution, where phase space is the six-dimensional space of position x and momentum p. What is the continuity equation for a phase space distribution? Consider a parcel of photons, spreading over a hyper-volume in phase space. As time passes from $t$ to $t^{\prime}$, the parcel moves from (x,p) to $(x^{\prime},p^{\prime})$. \begin{equation} {f(t,x,p)d^3 x d^3 p} = {f(t^{\prime},x^{\prime},p^{\prime})d^3 x^{\prime} d^3 p^{\prime}} \end{equation} According to Liouville's theorem, the phase space volume occupied by the parcel remains fixed: $d^3xd^3p=d^3x^{\prime}d^3p^{\prime}$ (See Figure below), and thus we must have $f(t,x,p)=f(t^{\prime},x^{\prime},p^{\prime})$! This means that the total time derivative of f must vanish; this gives the continuity equation of phase space distribution f: \begin{equation} \label{morepartials} {{1 \over c}{\partial \over \partial t}{I_{\nu}} + \hat \Omega \cdot \nabla I_{\nu} + {1 \over c}{\vec{\dot{p}} \cdot \nabla_p I_{\nu}}} = 0, \end{equation} where the last term represents processes such as gravitational redshift (photon energy momentum $E=pc=h\nu$). \begin{figure}[!h] \centering \includegraphics[width=\textwidth]{PhaseSpaceVol_GSnotes.png} \caption{As a parcel of photons moves through phase space, it's volume remains unchanged. \label{phasespacevol}} \end{figure} \end{itemize} \section{Application to 21cm Global Signal} \begin{itemize} \item We want to apply this to the 21cm brightness temperature, $T_B$. But how? \item Hydrogen's fine structure: see Figure below \begin{figure}[!h] \centering \includegraphics[width=\textwidth]{HfineStruc_GSnotes.jpg} \caption{A simple illustration of hydrogen's fine structure. Note that the n=1 orbital exhibits hyperfine splitting, which can result in a 21cm photon being emitted or absorbed depending on the direction of transition between the levels (from high to low, emission; from low to high, absorption). \label{Hfinestruc}} \end{figure} \item To understand the interaction of hydrogen with photons, we need to understand the relative population of these levels. In \underline{statistical equilibrium}, the relative population of two levels is given by the Boltzmann Distribution: \\ \begin{equation} \text{2 levels: 0, 1} \end{equation} \begin{equation} {n_1 \over n_0} = { {g_1 \over g_0} e^{- \Delta E / kT} } \end{equation}

\item {\bf Digression:} We now digress to give a motivation for the famous Boltzmann Distribution: \\ First, phase space density is solely a function of energy in statistical equilibrium: \begin{equation} f(x,p) = f(E) \end{equation} Why is this true? That is, why is the equilibrium phase space distribution a function of only energy $f(x,p,t) = f(E)$? \\ Classically, we have \begin{equation} {df \over dt} = {\partial f \over \partial t} + {\dot{x} \cdot \nabla f} + {\dot{p} \cdot \nabla_p f} \end{equation} \begin{equation} {df \over dt} = {\partial f \over \partial t} + \Bigg[ { {{\partial H} \over {\partial p}}{{\partial f} \over {\partial x}} - {{\partial H} \over {\partial x}}{{\partial f} \over {\partial p}}} \Bigg] \end{equation} where H is the Hamiltonian. Writing the second term on the right hand side as the Poisson Bracket [H,f], we have: \begin{equation} {df \over dt} = {\partial f \over \partial t} + [H,f] = 0 \end{equation} From this, we can see that \begin{equation} {{\partial f} \over {\partial t}} = - [H,f] = 0 \end{equation} must be true universally, i.e. for all forms of H. So f must only be a function of $H (=E)$.

Quantum mechanically, the Poisson bracket becomes the commutator. \begin{equation} {{\partial f} \over {\partial t}} = \frac{h}{i}[H,f] = 0 \end{equation} However the careful reader would realize that the phase space distribution has no simple generalization to the quantum picture, for the uncertainty relations prevent a phase space point from being well-defined. Although we shall not go into this here, one consistent generalization of the density matrix to phase space was given by Wigner. \item Gedanken Experiment: \\ Now consider 2 non-interacting systems: \begin{equation} \text{System 1: } E_1 \end{equation} \begin{equation} \text{System 2: } E_2 \end{equation} \begin{equation} \text{Total: } E_{12} = E_1 + E_2 \end{equation} \begin{equation} f_{12} = f_1(E_1)f_2(E_2) = f(E_1 + E_2) \end{equation} \begin{equation} \text{the set of functions f that satisfy this property are the exponentials:} \end{equation} \begin{equation} {e^{-\beta E_1} e^{-\beta E_2}} = {e^{-\beta (E_1 + E_2)}} \text{, where } {\beta \equiv {1 \over kT}} \end{equation} \item Thus all micro-states with a certain energy have the same weight of sampling, and thus the same population. When we want to group multiple micro-states together, we use the multiplicity factor g. For example, the figure above shows the lowest atomic orbitals of neutral hydrogen, and the figure below} shows the probability of obtaining a particular n (in phase space) given x and vice versa. If we group the micro-states by principal quantum number, we have \begin{figure}[!h] \centering \includegraphics[width=\textwidth]{prob_GSnotes.png} \caption{Illustration of the finding the probability of getting a particular n (in phase space) given x and vice versa. \label{prob}} \end{figure} \begin{equation} {g(n=2)} = 6 \text{ micro-states possible} \end{equation} \begin{equation} {g(n=1)} = 2 \text{ micro-states possible} \end{equation} \item And so ends our digression. And now, back to the 21cm global signal. We define a series of temperatures: \\ \underline{Spin temp:} \begin{equation}

{n_1 \over n_0} = { {g_1 \over g_0}e^{ {-\Delta E_{21cm} \over kT_{spin}}}}

\end{equation} \underline{CMB temp (blackbody):} \begin{equation} I_{\nu} = {{2h \nu^3} \over c^2}{1 \over {e^{h\nu / kT_{CMB}} - 1}} \end{equation} We define the CMB temperature such that the blackbody form is retained. It does tell you the distribution and average energy, but it's not exactly a \underline{real} temperature. Practically speaking though, it's useful to define something as a temperature. Note, this is NOT in LTE!!! \\ \\ \underline{$T_K$:} the kinetic temperature of the gas $\rightarrow$ this is a real temperature! \\ \\ \underline{$T_b$:} brightness temperature. Not a real temperature but defined as the temperature needed to obtain a measured intensity if the source is a blackbody: \begin{equation} I_{\nu} = {2k_B \over \lambda_{21cm}^2}T_b \end{equation} \item As we saw in the last talk (Talk 2), there are various temperatures involved in the evolution of the 21 cm hydrogen signal. The figure below shows the three temperatures we've previously discussed and how they are coupled and adds the color temperature -- the characteristic temperature of the Wouthuysen-Field effect. \begin{figure}[!h] \centering \includegraphics[width=\textwidth]{Coupling_GSnotes.png} \caption{Diagram of the various temperatures involved in the evolution of the 21 cm hydrogen signal. This is similar to the diagram in Talk 2, but has now added the color temperature -- the characteristic temperature of the Wouthuysen-Field effect. \label{coupling}} \end{figure} \item There are many different possible shapes to the 21 cm signal's evolution graph (see Figure below) depending on what actually happened and when. \begin{figure}[!h] \centering \includegraphics[width=\textwidth]{21cmsig_GSnotes.png} \caption{Plot of the global 21 cm signal's evolution for a fiducial model in differential brightness temperature versus frequency. Note that the evolution of the signal depends on what happened and when. \label{21cmsig}} \end{figure} \item From ${dI_{\nu} \over d\tau} = -I_{\nu} + S_{\nu}$, we can approximate $S_{\nu}$ as a blackbody: \begin{equation} S_{\nu} = { {2h\nu_{21cm}^3 \over c^2}{1 \over {e^{h\nu_{21cm} / kT_{spin}} - 1}} } \end{equation} \item We often find it helpful to define: \begin{equation} T_* = { {h \nu_{21cm}} \over k } = 0.068K << T_s,T_K, etc. \end{equation} \item We then have \begin{equation} S_{\nu} = { {2h\nu^3 \over c^2}{1 \over {e^{T_* / T_{spin}} - 1}} } \end{equation} \item We can then Taylor expand and use the Rayleigh-Jeans limit to obtain \begin{equation} S_{\nu} = {2h\nu^3 \over c^2}{T_{spin} \over T_*} = {2kT_{spin} \over \lambda_{21cm}^2} \end{equation} \item Recall $I_{\nu} = {2kT_b \over \lambda^2}$, which is exactly the same form as the previous equation. From this, we can rewrite ${dI_{\nu} \over d\tau} = -I_{\nu} + S_{\nu}$ in terms of the temperatures. Dividing out all the shared factors, we have \begin{equation} {dT_b \over d\tau} = -T_b + T_{spin} \end{equation} \item To solve this, we need an initial condition: $T_b(\tau = 0) = T_{CMB}$. That is, before we have anything else, we have the CMB. We then have as a solution for the differential brightness temperature: \begin{equation} {\delta T_b} = {T_b - T_{CMB}} = { {({T_{spin} - T_{CMB}}) \over ({1+z})}{(1-e^{-t})} } \end{equation} where the $(1+z)$ is a correction factor, accounting for the expansion of the universe. \item If $\delta T_b$ is positive, negative, or zero, we have emission, absorption, or no signal, respectively. \item LTE is reached when $T_K = T_{spin} = T_b$ \item Collisions set the rate of moving from one energy level to the other. \item For the $n=1$ levels (let the lower energy level be 0 and the higher level 1), let the rates of transition between the levels be $C_{01}$ (from 0 to 1) and $C_{10}$ (from 1 to 0). \item If in equilibrium (i.e. if collisions set the spin temperature), \begin{equation} n_0C_{01} = n_1C_{10} \end{equation} \item We can solve for the ratio of the rates \begin{equation} {C_{01} \over C_{10}} = {n_1 \over n_0} = { {g_1 \over g_0}{e^{-T_* / T_K}} } \end{equation} where the last equation holds outside of LTE as well and ${g_1 \over g_0} = 3$ (due to the spin triplet state). \item Taylor expanding, we have \begin{equation} {C_{01} \over C_{10}} \approx 3 \Bigg({1-{T_* \over T_K}} \Bigg) \end{equation} \item Similarly, the Wouthuysen-Field effect sets the rate of transition: \begin{equation} {W_{01} \over W_{10}} \approx 3 \Bigg({1-{T_* \over T_w}} \Bigg) \end{equation} \item To get all the astrophysical processes at play, we must write a balance equation: \begin{equation} \label{bala} n_1({A_{10} + B_{10}I_{CMB} + C_{10} + W_{10}}) = n_0({B_{01}I_{CMB} + C_{01} + W_{01}}) \end{equation} \item $A_{10}$, $B_{10}$, and $B_{01}$ are Einstein coefficients/rates: \begin{equation} A_{10} = \text{rate of spontaneous decay} \end{equation} \begin{equation} B_{10} = \text{stimulated emission} \end{equation} \begin{equation} B_{01} = \text{absorption} \end{equation} \item In the absence of $C$'s and $W$'s, the above simplifies to \begin{equation} n_1({A_{10} + B_{10}I_\nu}) = n_0({B_{01}I_\nu}) \end{equation} If equilibrium is reached, $I_\nu$ must be a black body: \begin{equation} I_\nu = I(A,B) = { {2h{\nu'}^3 \over c^2}{1 \over {e^{h\nu / kT} - 1}} } \end{equation} This gives us equilibrium ratios of the Einstein coefficients. (If you are not familiar with Einstein coefficients and their equilibrium ratios, refer to Rybicki and Lightman (RL) and/or AstroBaki.) \item From the above discussions, we obtain from the balance equation: \begin{equation} {n_1 \over n_0} = { {g_1 \over g_0}{e^{-T_{spin} / T_*}} } \approx {1 - {T_{spin} \over T_*}} \end{equation} We then obtain: \begin{equation} T_{spin}^{-1} = { {T_{CMB}^{-1} + x_cT_K^{-1} + x_wT_w^{-1}} \over {1 + x_c + x_w} } \end{equation} where \begin{equation} x_c = {C_{10} \over A_{10}}{T_* \over T_{CMB}} \end{equation} and \begin{equation} x_w = {W_{10} \over A_{10}}{T_* \over T_{CMB}} \end{equation} \item Therefore, $T_{spin}$ depends on all these different processes, which have different rates. See Figure below. \begin{figure} \centering \includegraphics[width=\textwidth]{deltaTb_GSnotes.png} \caption{Illustration of the various stages of the global 21 cm signal's evolution and its dependence on temperature and the energy level transition rate. Recall that $\delta T_b = T_{spin} - T_{CMB}$, $T_{CMB} \propto a^{-1}$, and $T_{gas} \propto a^{-2}$. \label{deltaTb}} \end{figure} \end{itemize}

\section{Wouthuysen-Field Effect} The WF effect is the process for a hydrogen atom to absorb a Lyman alpha photon, exciting the atom into an n=2 state. If it then decays back into a different n=1 hyperfine state, the spin flips in the process, producing a 21 cm photon. The WF effect becomes important during reionization due to the abundance of Lyman-alpha photons. The WF effect is also relatively competitive due to the quantum mechanical selection rules. Laws such as the conservation of parity and angular momentum tell us that only certain transitions are allowed. In particular, the electric dipole mode (strongest) of the direct 21cm transition is forbidden by parity, thus making it a factor $\alpha^2$ weaker.

The WF effect is discussed in detail on astrobaki here: Wouthuysen Field effect