# Talk 2: Differential Brightness Temperature and the Global 21cm Signal Evolution

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\title{Adrian Liu's Summer 2015 Global Signal Talks: Part 2: Evolution of the Global 21cm Signal} \author{Notes by Cherie Day} \date{6/18/2015}

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\section{Background} \begin{itemize}

\item Global Signal $\equiv$ average sky over whole sky (all angles); plot result as a function of frequency $\rightarrow$ $I(\nu)$ [or $I(\lambda)$] -- a spectrum! \item Zeroth order of cosmological signal takes form of blackbody spectrum of CMB. See Figure below. \begin{figure} \centering \includegraphics[width=\textwidth]{BBspec_intense_GSnotes.png} \includegraphics[width=\textwidth]{BBspec_tempunits_GSnotes.png} \caption{Zeroth order of cosmological signal has the form of a blackbody, and the global 21cm signal lies on top of the CMB signal. The top figure is the regular form of the spectrum (intensity versus frequency), and the bottom figure is the spectrum in units of temperature (still intensity versus frequency). Note in the latter, the spectrum is flat (uniform constant). \label{BBspec}} \end{figure} \item The CMB spectrum is fragile $\rightarrow$ anything in the way can cause deviations. \item Hydrogen can absorb or emit 21cm (1420MHz) wavelength photons due to the spin flip" or hyperfine transition, see next Figure. \begin{figure} \centering \includegraphics[width=\textwidth]{hyperfinetrans_GSnotes.png} \caption{Illustration of the hyperfine transition in neutral hydrogen. A 21cm wavelength photon can be emitted or absorbed depending on the state to which it's transitioning. \label{hyper}} \end{figure} \item As photons travel to us, the universe expands and wavelengths get stretched out. This can be seen from: $$E_{photon} \sim {k_B T} \sim {h \nu} \sim {1 \over \lambda}$$ where $k_BT$ is the average energy of a photon, $k_B$ is Boltzmann's constant, and $h$ is the Planck constant. Since $\lambda \propto a(t)$ -- the scale factor of our universe -- we find that $$T_{CMB} \propto {1 \over a(t)}$$ \item A CMB photon can interact with neutral hydrogen in a gas cloud on its way to us and result in a 21cm wavelength photon. Depending on where the cloud is along our line of sight, the wavelength can be stretched greatly (if it's far from us) or not very much (if it's very close). See Figure below. \begin{figure} \centering \includegraphics[width=\textwidth]{PhotonStretching_GSnotes.png} \caption{Two scenarios of CMB photons interacting with a cloud of hydrogen gas placed at different distances along the line of sight. With longer travel times post-interaction, the wavelengths are stretched more, and with shorter travel times, they are closer to 21cm. \label{stretching}} \end{figure} \begin{itemize} \item Long $\lambda$ $\rightarrow$ lower frequency: Must originate from hydrogen gas far away \\ $\Rightarrow$ early times in cosmic history. \item Short $\lambda$ $\rightarrow$ higher frequency: Must originate from hydrogen gas close by \\ $\Rightarrow$ late times in the universe's history (i.e. emitted more recently) \item See Figure below for how time relates to frequency. \begin{figure} \centering \includegraphics[width=\textwidth]{CosmoTime_GSnotes.png} \caption{Time-frequency relation. Higher frequency $\Rightarrow$ more recent. Lower frequency $\Rightarrow$ earlier times in cosmic history. \label{cosmictime}} \end{figure} \end{itemize} \item $T_{spin}$ $\rightarrow$ Measure of the relative number of hydrogen atoms in the excited versus ground spin flip" state. \item High $T_{spin}$ $\rightarrow$ Mostly in excited spin state (proton and electron spins aligned) \\ $\rightarrow$ Likely to emit 21cm photons \\ $\rightarrow$ e.g. If $T_{spin} \sim \infty$, the only option it has is to emit. Thus, the more atoms there are in the excited state, the more likely they are to emit. \item Low $T_{spin}$ $\rightarrow$ Mostly in ground state (proton and electron spins anti-aligned) \\ $\rightarrow$ Likely to absorb 21cm photons

\end{itemize}

\section{Differential Brightness Temperature} \begin{itemize} \item The differential brightness temperature is an expression quantifying the difference of the brightness temperature (obtained from the radiative transfer equation [See Figure below for an illustration]) and the CMB temperature $\Rightarrow$ $\delta T_b = T_b - T_{CMB}$. This highlights that when the brightness temperature differs from the $T_{CMB}$, it becomes meaningful. \begin{figure}[!h] \centering \includegraphics[width=\textwidth]{radtrans_GSnotes.png} \caption{Illustration of components relavant to radiative transfer. CMB photons travel towards us, and on their way, they interact with a cloud of hydrogen atoms with temperature $T_{spin}$, resulting in the measured brightness temperature, $T_b$. \label{radtrans}} \end{figure} \item The differential brightness temperature for $T_{spin}$ relative to $T_{CMB}$ can be written as the following proportionality $${\delta T_b} \propto { ({1 + \delta_b}) x_{HI} ({1 - {T_{CMB} \over T_{spin}}}) }$$

\begin{itemize}

\item $x_{HI}$: neutral fraction of hydrogen. It ranges from 0 to 1 \\ $\rightarrow$ the signal we see is proportional to $x_{HI}$ because the transition occurs only with neutral hydrogen \item $\delta_b$: fractional overdensity in baryons (i.e. the amount it is denser than another part of the universe) \\ $\rightarrow$ $1 + \delta_b =$ density \item $\delta T_b \propto ({1 - {T_{CMB} \over T_{spin}}})$ and we can discuss various limiting cases:

\begin{enumerate} \item $T_{spin} >> T_{CMB}$ $\Rightarrow$ $\delta T_b$ is positive \\ $\rightarrow$ more emission happening and so $T_{spin}$ becomes larger than $T_{CMB}$ $\rightarrow$ Note: a maximum can be obtained of $+1$ \item $T_{spin} = T_{CMB}$ $\Rightarrow$ $\delta T_b = 0$ \\ $\rightarrow$ No contrast because it's just as likely a photon will be absorbed as emitted since the two are in equilibrium. \item $T_{spin} << T_{CMB}$ $\Rightarrow$ $\delta T_b$ is negative \\ $\rightarrow$ Mostly absorbing $\rightarrow$ Note: there is no bound on how negative this can be unlike in the emission case. \end{enumerate}

\end{itemize} \end{itemize}

\section{Phases of the Global 21cm Signal's Evolution}

\indent There are a few temperatures involved here, and we must explain how these evolve over time to see why the global signal graph we've been drawing (e.g. in time-frequency relation figure above) looks as it does. $T_{spin}$ is the trickiest. The figure below shows schematically the temperatures and processes involved in the phases leading up to and including reionization. \begin{figure} \centering \includegraphics[width=\textwidth]{temps_GSnotes.png} \caption{Schematic of the temperatures involved in the various phases of the Global 21cm signal's evolution. Note that the gas and CMB temperatures are actual temperatures while the spin temperature measures the relative number of excited versus ground hyperfine state hydrogen atoms. \label{temps}} \end{figure}

Phase I: Initially no signal \begin{itemize} \item Residual electrons drive $T_{Gas}$ to $T_{CMB}$ via Compton scattering $\Rightarrow$ photons hit electrons $\rightarrow$ electrons collide with gas, heating it \item Collisions drive $T_{Gas}$ to $T_{spin}$ $\Rightarrow$ the gas collides with itself \item If the coupling is strong enough, the temperatures come into equilibrium. Thus, we can't go to arbitrarily early times because the signal disappears. \end{itemize}

Phase II: Absorption \begin{itemize} \item Collisions maintain $T_{Gas}$ and $T_{spin}$ equilibrium \item CMB has become too dilute for $T_{CMB}$ and $T_{Gas}$ to remain coupled $\rightarrow$ the gas can do what it wants now \itemWe also have: $$T_{spin} = {T_{gas} \sim a^{-2}}$$ while $$T_{CMB} \sim a^{-1}$$ \item Why is the last statement true? If you have adiabatic expansion (as is true for our universe), then $$pV^{\gamma} = constant$$ For a monatomic gas like hydrogen: $\gamma = {5 \over 3}$ Using the Ideal Gas Law, this can being rewritten as $$TV^{\gamma-1} = constant$$ So for us (with hydrogen), we have $$TV^{2/3} = constant$$ This is just expanding with the universe (volume expands), so given $V \propto a^3$ and that $T = T_{Gas}$, we have $$T_{Gas}a^2 = constant \Rightarrow T_{Gas} \propto {1 \over a^2}$$ Recall from Talk 1 that $T_{CMB} \propto a^{-1}$, so the gas is cooling faster than the CMB! \item Since collisions keep $T_{Gas}$ and $T_{spin}$ coupled, the gas is a net absorber of 21cm radiation. \end{itemize}

Phase III: Back to nothing \begin{itemize} \item Gas becomes too dilute for collisions to keep $T_{Gas}$ and $T_{spin}$ in equilibrium \item Absorption and emission of 21cm photons makes $T_{spin} = T_{CMB}$ \\ $\rightarrow$ no net absorption or emission \end{itemize}

Phase IV: First Stars and Galaxies Form \begin{itemize} \item First stars create $Ly \alpha$ photons which have the right energy to cause the hyperfine transition. i.e. $Ly \alpha$ photons are perfect for causing the absorption and emission of 21cm radiation. \\ $\rightarrow$ This causes coupling between $T_{Gas}$ and $T_{spin}$ via the Wouthuysen-Field effect (to be discussed in the next installment). \item $T_{spin} = T_{Gas} < T_{CMB}$ $\Rightarrow$ absorption \end{itemize}

Phase V: X-ray reheating \begin{itemize} \item X-rays from early AGNs (Active Galactic Nuclei) heat the IGM (Intergalactic Medium), raising $T_{Gas}$ which is still coupled to $T_{spin}$ because of the Wouthuysen-Field effect. \item Electrons undergo all sorts of atomic transitions, giving off all sorts of photons causing $$T_{spin} = T_{Gas} > T_{CMB} \Rightarrow \text{emission}$$ \end{itemize}

Phase VI: Reionization \begin{itemize} \item Eventually, reionization rids us of all the neutral hydrogen atoms. The stars and galaxies are pumping UV photons into the IGM, and these have a resonant energy for ionizing hydrogen. i.e. They have the highest probability of ionizing atomic hydrogen: while other photon frequencies have the energy to ionize, they have a lower probability of ionizing than do UV photons, so other frequencies (like X-rays) are less efficient. \item No more signal! Since the signal is produced only by atomic (neutral) hydrogen, the signal disappears when the neutral hydrogen disappears. \end{itemize}

\noindent If you don't do the average (getting one number per frequency), you get a lot of rich information. There are finer structures in the actual sky. Note also that this is a fiducial model of the global signal's evolution, which is based in physics but has not been verified $\rightarrow$ no one really knows with certainty what happened. See Figure below.

\begin{figure} \centering \includegraphics[width=\textwidth]{PritchardLoeb2012_GlobalSigEvo_GSnotes.png} \caption{Fiducial model of the 21cm cosmic hydrogen signal from Pritchard and Loeb 2012 (http://arxiv.org/pdf/1109.6012.pdf). The top figure shows the fluctuations in the 21cm brightness temperature from before the first luminous sources through reionization. It's evident, as discussed in the text, that the non-averaged signal contains fine structure and rich information. The bottom figure is the fiducial model of the evolution of the global (sky-averaged) 21cm signal (solid curve). The dashed line shows $T_b = 0$. This graph also illustrates that with the 21cm hydrogen signal, when we observe at different frequencies, we're observing the signal at different redshifts. See Pritchard and Loeb 2012 for more on how these figures were made. \label{globalsigevo}} \end{figure}

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