SZ Effect

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\section*{ The Sunyaev-Zeldovich Effect: A case-study of Inverse-Compton Scattering}

A classic astrophysical example of Inverse Compton Scattering is the Sunyaev-Zel'dovich (SZ) effect.

Massive galaxy clusters throughout the Universe are filled with fast-moving free electrons. Most CMB photons have been free-streaming through the Universe since the Epoch of Recombination. However, a fraction of CMB photons pass through massive clusters before reaching our detector, and some of these will interact with high energy electrons through inverse Compton scattering. This changes the energies of the CMB photons and produces an observable deviation from the Planck function in the CMB spectrum.

To see quantitatively how interactions with energetic photons change the CMB spectrum, let's consider what happens to a single photon passing through a cluster, which for our purposes can be approximated as a cloud of hot electrons.

We'll call the kinetic temperature of electrons $T_{e}$. As derived in the lecture on Inverse Compton Scattering, the power radiated by a single electron passing through a photon field is $$P=\frac{4}{3}\sigma_{T}U_{\gamma}c\beta^{2}\gamma^{2},$$ where $\sigma_{T}$ is the Thomson cross section, $U_{\gamma}$ is the energy density of the radiation field, $\beta=v/c$, and $\gamma = 1/\sqrt{1-v^2/c^2}$.

We can calculate $\Delta E_{\gamma}$, the typical change in the energy of a photon for a single interaction with an electron, as $\Delta E_{\gamma}=P/R_Template:\rm coll$, where $R_Template:\rm coll=N_Template:\rm coll/\Delta t$ is is the number of collisions with photons per unit time for a single electron.

The typical number of collisions can be written as $\tau=n\sigma s$, and $\Delta t=s/v$.

Thus, $$R_Template:\rm coll=\frac{N_Template:\rm coll}{\Delta t}=\frac{n_{\gamma}\sigma_{T}s}{s/v}=n_{\gamma}\sigma_{T}v$$

Here $n_{\gamma}$ is the number density of photons, $\sigma_{T}$ is the Thomson cross section for electron-photon interactions, and $v$ is the typical electron velocity. We can then write $$\Delta E_{\gamma}=\frac{P}{R_Template:\rm coll}=\frac{\frac{4}{3}\sigma_{T}U_{\gamma}c\beta^{2}\gamma^{2}}{n_{\gamma}\sigma_{T}v}=\frac{4}{3}\gamma^{2}\beta^{2}E_{\gamma}$$


Where we have introduce $E_{\gamma}=U_{\gamma}/n_{\gamma}$, the average photon energy.

We'd like to write $\Delta E_{\gamma}$ in terms of temperature. From energy equipartition, we have (for a single electron): $$\frac{1}{2}m_{e}v^{2}=\frac{3}{2}kT_{e}\Rightarrow v^{2}=\frac{3kT_{e}}{m_{e}}.$$

We're going to assume that $v\ll c$, so that $\gamma\to1$.

Then we can write $$\Delta E_{\gamma}\approx\frac{4}{3}\beta^{2}E_{\gamma}=\frac{4v^{2}}{3c^{2}}E_{\gamma}=\frac{4kT_{e}}{m_{e}c^{2}}E_{\gamma}$$

This expression gives the change in the energy of a photon after a single interaction with an electron. If a 2nd interaction occurs, the electron's energy will change again. In the limit of many interactions $N$, we can change the discrete $\Delta E$ into an infinitesimal differential, and we have a differential equation: $$\frac{{\rm d}E_{\gamma}}{{\rm d}N}=\left(4\frac{kT_{e}}{m_{e}c^{2}}N\right)E_{\gamma}=4yE_{\gamma}$$

where we introduced the dimensionless Compton $y$ parameter, $y\equiv kT_{e}N/\left(m_{e}c^{2}\right)$.


This is a homogenous differential equation for $E_{\gamma}$; it's solution is $$E_{\gamma}=E_{\gamma,0}e^{4y}$$

For a typical cluster, $n_{e}\approx3\times10^{-3}\,{\rm cm}^{-3}$ and $\ell=3\times10^{24}\,{\rm cm}.$ Thus, the Optical Depth is $$\tau_{e}=n_{e}\sigma_{T}\ell\approx10^{-2}$$

I.e., clusters are optically thin. Thus, we can write $N\approx\tau$. For a cluster with electron temperature $T_{e}=10^{8}{\rm K}$ (corresponding to a mass of a few $\times10^{14}\,M_{\odot}$), we have $$\frac{kT_{e}}{m_{e}c^{2}}\approx10^{-2};\qquad y=10^{-2}\times10^{-2}=10^{-4}$$


The fact that $\tau_{e}\ll1$ means that most CMB photons passing through a cluster will not interact at all. Of the approximately $1\%$ that do, almost all will only interact once, and their energies will typically increase by a factor of $\sim e^{4\times10^{-4}}\sim1+4\times10^{-4}$.

\section*{Effect on CMB spectrum}

How does the SZ Effect change the CMB spectrum in practice? Figure 1 compares the predicted CMB spectrum after the SZ Effect (solid line) to the blackbody spectrum predicted without the SZ effect (dashed line). Since the energies of CMB photons are much lower than cluster electrons, all CMB photons which interact with electrons gain energy. However, because the total number of photons in conserved, the SZ effect decreases the number of photons at lower energies, moving the to higher energy regions of the spectrum.

Effect of SZ effect on the CMB spectrum. The dashed line shows the expected spectrum for a blackbody of with the temperature of the CMB. The solid line shows the spectrum after the SZ effect: photons are removed from the lower-energy part of the spectrum and upscattered to higher energies. Adapted from Figure 1 of Carlstrom et al., Annual Reviews of Astronomy & Astrophysics vol 40, pg 643, 2002.

This can be seen more clearly in Figure 2, below, which plots the difference between the CMB spectrum after the SZ effect and the pure blackbody spectrum. The number of photons with frequencies less than 218 GHz is decreased, while the number with energies greater than 218 GHz is increased.

Difference between a blackbody spectrum and the post-SZ effect CMB spectrum. The number of photons with frequencies less than 218 GHz is decreased, while the number with energies greater than 218 GHz is increased. Data from the cluster Abell 2163. Adapted from Figure 4 of Carlstrom et al., Annual Reviews of Astronomy & Astrophysics vol 40, pg 643, 2002.

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\iffalse Consider a non-relativistic thermal bath of electrons at temperature $T_e$. The average kinetic energy of these electrons is ${3\over2}kT_e$. Now suppose that there is also a bath of photons which all have energy $E=h\nu\ll kT_e$. Putting this cold bath of photons in with the hot (but non-relativistic) electrons, we find that they get up-scattered. After first scattering, the mean shift in energy is: \begin{aligned}\Delta\_E&={P_{K,single\ e^-}\over n_{photons}\sigma_Tc}= {{4\over3}U_{ph}\gamma^2\beta^2\sigma_Tc\over n_{photons}\sigma_Tc}= {4\over3}\gamma^2\beta^2E\\ {\Delta\_E\over E}&={4\over3}\gamma^2\beta^2={4\over3}{v^2\over c^2}= {4\over3}{3kT_e\over m_ec^2}={4kT_e\over m_ec^2}\\ \end{aligned} which is much the mean fractional change in photon energy after 1 scattering.

Last time, we'd written down the mean energy of the upscattered photons: $$\Delta\_E={P_{Compton,single\ e^-}\over n_{photons}\sigma_Tc}$$ The bottom term $n_{photons}\sigma_Tc$ is just the collision rate for photons. We used this to calculate: $${\Delta\_E\over E}={4kT_e\over m_ec^2}\ll1$$ Thus, after a single scattering: $$\_E_i=E\left(1+{4kT_e\over m_ec^2}\right)$$ So after $N$ scatterings: $$\_E_N=E\left(1+{4kT_e\over m_ec^2}\right)^N$$ In the limit of $N\to\infty$: $$E_N=Ee^{4kT_eN\over m_ec^2}\equiv Ee^{4y}$$ This is the definition of the Compton $y$ parameter: $y\equiv {kT_e\over m_e c^2}N$.\par Suppose we have a cloud of electrons that has radius $l$, and the [[Optical Depth]] out of the cloud $\tau_e\gg1$. The mean number of collisions a photon will undergo in getting out of the cloud is $N\sim\tau_e^2$. This is because each time step between collisions is ${\lambda_{mfp}\over c}$, and the number of steps to get out is: $${Time\ to\ get\ out\over Each\ time\ step}\sim{{l^2\over c\lambda_{mfp}} \over {\lambda_{mfp}\over c}}\sim\left({l\over\lambda_{mfp}}\right)^2 \sim\tau_e^2$$ \def\ppt#1Template:\partial The reason why the time to get out is $l^2\over c\lambda_{mfp}$, is because the diffusion equation gives us $\ppt{Q}=D\nabla^2Q$, which by dimensional analysis, says: $$\inv{T}\sim{D\over L^2}$$ $D$ is the diffusivity (viscosity) of the medium.\par If $\tau_e\ll1$, $N_s\sim\tau$ because the intensity goes as $e^{-\tau}\approx 1-\tau$. Generally: \begin{aligned}N_s&\sim max(\tau_e,\tau_e^2)\\ &\sim\tau_e(1+\tau_e)\\ \end{aligned} Thus, the Compton $y$ parameter is: $$y\approx {kT_e\over m_ec^2}\tau_e(1+\tau_e)$$ \begin{itemize} \item We'll apply this to CMB photons traveling through an intra-cluster gas. We'll say: $n_e\sim3\e{-3}cm^{-3}$, $\sigma_T\sim10^{-24}cm^2$, and $l\sim10^6\cdot3\e{18}cm$. This gives us $\tau_e\sim10^{-2}$. Now $kT_e$ should be a few $keV$, so $${kT_e\over m_ec^2}\sim10^{-2}$$ The us the energy post-traversal of the cluster is related to the initial energy of the photons by: $$E_{post}=E_{init}e^{4y}=E_{init}(1+4\e{-4})$$ Thus, we expect to see a shift in the frequency of peak flux when looking through a cluster vs. around one. We can calculate the change in brightness temperature this causes by using: \begin{aligned}F_{\nu,before}&={2kT_{CMB}\over\lambda^2}\\ F_{\nu,after}\left(\lambda={\lambda_0\over1+4y}\right) &={2kT_{CMB}\over\lambda^2}\\ &={2kT_{CMB}^\prime\over\lambda_0^2}\\ \end{aligned} Solving for $T_{CMB}^\prime$: \begin{aligned}T_{CMB}^\prime&=T_{CMB}\left({\lambda\over\lambda_0}\right)^2\\ &=T_{CMB}\left({1\over1+4y}\right)^2\approx T_{CMB}(1-8y)\\ \end{aligned} So the change in brightness temperature is $\Delta T_{CMB}\sim8yT_{CMB}\sim 10mK$. This is hard to detect from ground-based telescopes, but this is a much greater effect than the inherent anisotropies of the CMB (which we've detected), so space-based telescopes like WMAP, if they have the angular resolution, should have already detected this. Yet they haven't.\par Now we've been a little sloppy. We've said that every photon from the CMB underwent a small shift in energy, but for an optically thin cloud like the one we've been discussing, only a few photons are ever scattered. We'll examine this on the Rayleigh-Jeans tail.\par For any $F_\nu$ incident upon our cloud, the $F_{\nu,after}$ which results from unscattered photons should be less (there are fewer photons). Then the flux from the upscattered photons gets added back in. We'll choose some frequency $\nu_0$ which scatters into $\tn=\nu_0\left(1+{4kT_e\over m_ec^2} \right)$, so that: \begin{aligned}F_{\nu,after}(\tn) &={2kT_{CMB}\over c^2}\tn^2-{2kT_{CMB}\over c^2}\tn^2\tau_e +{2kT_{CMB}\over c^2}\nu_0^2\tau_e\\ &={2kT_{CMB}\over c^2} {\tn^2\over\left(1+{4kT_e\tau_e\over m_ec^2}\right)^2}\\ \end{aligned} This matches what we estimated just by shifting, so we're okay.\par

\item We'll examine another situation where we can use the SZ effect to measure $H_0$ in an x-ray cluster. We'll say that $\theta_c$ is the observed angular size of the cluster, and $F_x$ is the observed x-ray flux. We'll say there's some observed redshift (using optical data), which is related to the recession velocity $v$. The x-ray luminosity $L_x= F_x\cdot4\pi d^2$, and the recessional velocity is $v=Hd$. Thus: $$L_x\propto F_xv^2H^{-2}$$ We also know, from bremsstrahlung that $L_x\propto n_e^2r_c^3F(T_e)$, where $F(T_e)$ is some function of the temperature of the electrons, and $r_c$ is the radius of the cluster. Now we can express $r_c$ as: $$r_c\sim\theta_cd\sim\theta_c{v\over H}$$ Setting our two expressions for the luminosity equal: \begin{aligned}F_xv^2H^{-2}&\propto n_e^2r_c^3F(T_e)\\ &\propto n_e^2\theta_c^3{v^3\over H^3}F(T_e)\\ \end{aligned} which gives us that $n_e\propto H^\hf$. Now SZ tells us that $y\propto T_en_e r_c\propto n_er_c\propto H^\hf\theta_c\cdot d{v\over H}$. Thus: $$y\propto \inv{H^\hf}$$ \end{itemize} \fi

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