Radiation Lecture 19

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<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} \def\^{\hat } \def\.{\dot } \def\tnTemplate:\tilde\nu

\usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document} \subsection*{ More Sunyaev-Zeldovich Effect}

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}

\subsection*{ Zeeman Effect}

The Zeeman Effect concerns the splitting of electronic levels in a magnetic field. We've already talked a little bit about this in hyperfine splitting, which was caused by magnetic fields intrinsic to an atom. We find that a single line can split into $\sim3-27$ components, depending on the number of combinations of $\vec L$ and $\vec S$ there are in the atom. The strengths of these various lines depend on viewing geometry, and can be polarized. The change in energy between previously degenerate states set up by an external $B$ field is: $$\begin{aligned}\Delta E&\sim\mu B\sim{e\hbar\over m_ec}B\\ &\sim{hc\over\lambda_1}-{hc\over\lambda_2}\\ &\sim{hc\over\lambda}{\Delta\lambda\over\lambda}\sim{e\hbar B\over m_ec}\\ \end{aligned}$$ Thus the fractional change in wavelength is: $$\boxed{{\Delta\lambda\over\lambda}\sim\lambda{eB\over2\pi m_ec^2}}$$ In practice, we find that the various split components of the original absorption lines are hard to resolve, and we see the effect expressed mostly as a broadening of the original line.

\end{document} <\latex>