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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 } \usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document} Recall that last time we derived for an optically thin synchrotron gas that: \def\pmag{{P_{mag}}} $$j_\nu(\nu_1)\propto P_eP_{mag}^{3\over4}\nu_1^{-\hf}$$ \begin{aligned}P_{tot}&=P_e+P_{mag}\\ &={Cj_\nu\over\pmag^{3\over4}\nu_1^{-\hf}}+\pmag\\ \end{aligned} Thus, the minimum total power occurs near the equipartition point between $\pmag$ and $P_e$. This gives us: \def\jn{j_\nu} $${C\jn\over\pmag^{3\over4}\nu_1^{-\hf}}\sim\pmag$$ $${B^2\over8\pi}=\pmag\sim(C\jn\nu_1^\hf)^{4\over7}$$
To discuss synchrotron self-absorption, we need to discuss what the spectrum of an optically {\it thick} medium of relativistic electrons looks like. If we take a bunch of $e^-$ spiraling around magnetic field lines with $\_E=\gamma m_ec^2$, then the energy of the photons emitted by these electrons is $h\nu\sim h\nu_{crit}\sim h\nu_{cyc}\gamma^2$. If this gas were optically thin, we'd just see a sharply peaked spectrum around $\nu_{crit}$ with $\nu\ll\nu_{crit}$ going as $\nu^{1\over3}$ and $\nu\gg\nu_{crit}$ going as $\nu^\hf e^{-{\nu\over\nu_{crit}}}$. Now let's add more electrons to this gas. For a while, the more electrons we add, the more emission we see. At some point, though, self-absorption starts making a difference, and we get less emission per electron added.\par Let's examine the peak amplitude of emission (that is, $I_\nu$ at $\nu=\nu_{crit}$). To do this, imagine we have an optically thick ball of blackbody (perfectly absorbing and emitting) particles with a temperature carefully chosen so that $kT\sim\gamma m_e c^2$. Then: \def\nucrit{{\nu_{crit}}} $$I_\nu(\nu=\nucrit)=B_\nu(T\sim{\gamma m_ec^2\over k},\nu=\nucrit)$$ Now we'd like to argue that this blackbody particle system is emissively identical to the optically thick synchrotron gas. To see why this is true, imagine we overlaid our sphere of blackbody particles on our sphere of synchrotron gas. Since all particles in it have the same energy, then energy cannot be transfered between the two system by collisions. The only option for exchanging energy is through photons of energy $h\nucrit$. However, this cannot create a net flow of energy because (I don't know).\par If $h\nucrit\ll\gamma m_ec^2$, then: \begin{aligned}I_\nu(\nucrit)^{\e,thick} &=B_\nu(T\sim{\gamma m_ec^2\over k},\nu=\nucrit)\\ &={2kT\over\lambda_{crit}^2}={2\gamma m_ec^2\over\lambda_{crit}^2}\\ &=2\gamma m_e\nucrit^2\\ \end{aligned} Since $\nucrit\sim\nu_{cyc}\gamma^2\propto B\gamma^2$, we have that $\gamma\sim\left({\nucrit\over B}\right)^\hf$. Thus, for an optically thick gas: $$I_\nu\propto\nucrit^{5\over2}B^{-\hf}\propto\nu^{5\over2}B^{-\hf}$$ It is important to remember that each $\gamma$ has a unique corresponding $\nucrit$. Also note that optical thickness depends inversely on frequency, so if we plot $I_\nu$ vs. $\nu$, we get a power law $\nu^{(1+p)\over2}$ for high frequencies (low optical thickness), and a $\nu^{5\over2}$ dependence for low frequencies (high optical thickness). There is also some $\nu_m$ where $I_\nu$ is maximal.
Compton scattering is the scattering of a photon off of an electron. If the photon {\it loses} energy, this is called {\bf Compton Scattering}. If the photon gains energy, this is called {\bf Inverse Compton Scattering}, or Compton Up-Scattering. The comptonization of an electron gas is the gain in energy of an electron gas as the result of a photon gas.\par Some examples of applications of Compton scattering are: \begin{itemize} \item Compton exchange keeps electrons in thermal equilibrium with photons at redshifts $z\ge10^3$. \item The spectra of AGN and xray binaries are altered by Compton Scattering (e.g. radio emission to optical wavelengths). \item CMB photons get upscattered by galaxy cluster plasma. This is called the Sunyaev-Zeldovich effect. \end{itemize} The classic model for Compton scattering is where a photon of energy $E=h\nu$ going in the $\^n$ direction scatters off of a stationary $e^-$. After scattering, the photon leaves with a new energy $E_1$ in a new direction $\^n_1$ (at an angle $\phi$ to its original direction $\^n$), and the $e^-$ now moves in direction $\^p$ with energy $E$ and momentum $p$. Energy conservation requires that: $$E+m_ec^2=E_1+E$$ and momentum conservation requires that: $${E\over c}\^n={E_1\over c}\^n+p\^p$$ Squaring these two equations and subtracting them, we find that: $$\lambda_1-\lambda={h\over m_ec}(1-\cos\phi)\leftrightarrow E_1={E\over1+{E\over m_ec^2}(1-\cos\phi)}$$ Comments: \begin{itemize} \item $\lambda_1-\lambda>0$: the shift here is tiny. The maximum possible value is $\lambda_1-\lambda={2h\over m_ec}=0.04\AA$. The momentum tends to be shared between the electron and photon, but no so much the energy. \item Remember this is {\it scattering}, not absorption. Photon \# is conserved. \item Note that for $h\nu\ll m_ec^2$, $\sigma=\sigma_T$. For $h\nu\gg m_ec^2$, $\sigma\sim\sigma_T\left({m_ec^2\over h\nu}\right)$. This additional term is called the Klein-Nishna correction. \item The mean scattering angle is $\phi={\pi\over 2}$. ${d\sigma_T\over d\Omega}\propto1+\cos^2\phi$. Beware that: $${d\sigma\over d\Omega}\eval{Klein-Nishna}\ne{d\sigma_T\over d\Omega}$$ \end{itemize} What we've done so far was for a stationary electron. For a moving electron, we need to consider the dependence on the angle at which the photon is coming in with respect to the direction of velocity ($\theta$). To make this situation similar to the one we just considered, we need to be in the frame of the electron. In this frame, the photons has a new energy as a result of time dilation: $$E^\prime=\gamma E(1-{v\over c}\cos\theta)$$ The $v\over c$ term is just the classic Doppler shift of the photon. Now suppose that the photon (which entered at angle $\theta^\prime$ in the electron frame) rebounds at an angle $\theta_1^\prime$. To relate this to our previous derivation, we want to find $\phi$. So note that $\theta_1^\prime-\phi^\prime=\theta^\prime$. Thus, in the electron's frame: $$E_1^\prime={E^\prime\over1+{E^\prime\over m_ec^2}(1-\cos\phi^\prime)}$$ Transforming this back into the lab frame: $$E_1=E_1^\prime\gamma(1+{v\over c}\cos\theta_1^\prime)$$ This generally follows Rybicki \& Lightman. However, it might help to know that in R\&L, 7.7b follows from 7.8a, which follows from 7.7a.\par Comments: \begin{itemize} \item If $E^\prime\ll m_ec^2$, then: \begin{aligned}E_1&\approx E^\prime\gamma(1+{v\over c}\cos\theta_1^\prime)\\ &\approx\gamma^2E(1+{v\over c}\cos\theta_1^\prime)(1-{v\over c}\cos\theta)\\ \end{aligned} In a typical collision, $\theta\sim\theta_1^\prime\sim{\pi\over2}$, so $E_1\sim\gamma^2E$. \item If $E^\prime\gg m_ec^2$, then: \begin{aligned}E_1^\prime&={E^\prime\over1+{E^\prime\over m_ec^2} (1-\cos\phi^\prime)}\approx m_ec^2\\ E_1&=E_1^\prime\gamma(1+{v\over c}\cos\theta_1^\prime)\\ &\approx m_ec^2\gamma(1+{v\over c}\cos\theta_1^\prime)\approx\gamma m_ec^2\\ \end{aligned} This final term defines the maximum rebound of the photon. \end{itemize}