<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} \def\numin{{\nu_{min}}} \def\numax{{\nu_{max}}} \def\gamin{\gamma_{min}} \def\gamax{\gamma_{max}} \def\gul{{min(\sqrt{\nu\over\numin},\gamax)}} \def\gll{{max(\sqrt{\nu\over\numax},\gamin)}} Recall in deriving the interaction of synchrotron radiation with synchrotron electrons, we derived the following formula for flux: $$F_{\nu,SSC}(\nu)=\tau\int_{\tn}{K\tn^\alpha d\tn\delta\left(\tn- {\nu\over\gamma^2}\right)\int_\gamma{N_0\gamma^sd\gamma}}$$ Now the integral over $\gamma$ is along slant paths through the rectangle ($\gamin\to\gamax$, $\numin\to\numax$). Some of these slant paths will not stretch all the way to $\gamax$ or $\gamin$ because of the boundaries imposed by $\numin$ and $\numax$. So we need to be a little more precise about the bounds on the $\gamma$ integral: \begin{aligned}F_{\nu,SSC}(\nu) &\sim\tau\int_\numin^\numax{K\tn^\alpha\delta(\tn-{\nu\over\gamma^2})d\tn \int_\gll^\gul{N_0\gamma^sd\gamma}}\\ &\sim\tau K\left({\nu\over\gamma^2}\right)^\alpha \int_\gll^\gul{N_0\gamma^sd\gamma}\\ &\sim\tau K\nu^\alpha N_0 \int_\gll^\gul{{\gamma^s\over\gamma^{2\alpha}}d\gamma}\\ \end{aligned} But since $\alpha={1+s\over2}$, the integrand is simply $\inv{\gamma}$, giving us: $$\boxed{F_{\nu,SSC}(\nu)\sim\tau k\nu^\alpha N_0\ln\left({\gul\over\gll}\right)}$$ Recall that $N_0$ was normalized so that $\int{N_0\gamma^sd\gamma}=1$, so saying that $\gamax$ is just some multiple of $\gamin$, it must be that $$N_0\sim\gamin^{-1-s}$$ Note for $\nu\sim\gamin^2\numin$, $F_\nu$ looks like (using the above relationship): \def\gllogul{\left({\gul\over\gll}\right)} \begin{aligned}F_{\nu,SSC}&=\tau K\gamin^{2\alpha}\numin^\alpha N_0\ln\gllogul\\ &=\tau K\numin^\alpha\gamin^{2\alpha}\gamin^{-1-s}\ln\gllogul\\ &=\tau K \numin^\alpha\ln\gllogul\\ \end{aligned}

\subsection*{ Compton Catastrophe}

If you keep scattering the same electrons, as in Synchrotron Self-Compton, there is a danger, if things are dense enough, of a runaway amplification of radiation energy density, or a Compton Cooling Catastrophe. However, we've never seen anything with a brightness temperature of $10^{12}K$. What sets this inverse Compton limit at this temperature? Comparing, for a single electron, the luminosity of inverse Compton scattering to synchrotron scattering: $${L_{IC}\over L_{sync}}={{4\over3}\beta^2\gamma^2\sigma_TcU_{ph}\over {4\over3}\beta^2\gamma^2\sigma_TcU_B}={U_{ph}\over U_B} \begin{cases} >1&catastrophe\\ <1 &no\ catastrophe\end{cases}$$ Now we're going to make an approximation that we are on the Rayleigh-Jeans side of the blackbody curve, so that: \begin{aligned}U_{ph}=U_{ph,sync}&\propto\nu_mI_\nu(\nu_m)\\ &\propto\nu_m{2kT_B\over\lambda_m^2}\\ &\propto\nu_m^3T_B\\ \end{aligned} where $\nu_m$ is the frequency of peak of synchrotron emission. Now $U_B\propto B^2$ is pretty obvious: $$\nu_m\sim\gamma_m^2\nu_{cyc}\propto\gamma_m^2B$$ where this $\gamma_m$ is not $\gamax$. Making the approximation that we are in the optically thick synchrotron spectrum, so that $\gamma m_ec^2\sim kT$, then we get $\nu_m\sim T_B^2B$. We can say that the kinetic temperature is the brightness temperature because we are talking about the average kinetic energy of the electrons generating the synchrotron radiation with a particular brightness temperature (i.e. another frequency of synchrotron radiation will have another brightness temperature, and another set of electrons moving with a different amount of kinetic energy). Thus, $${U_{ph}\over U_B}=C{\nu_m^3T_B\over\nu_m^2}T_B^4 =\left({\nu_m\over10^9Hz}\right)\left({T_B\over10^{12}K}\right)^5=1$$ A way of think about this is that, in order to avoid having infinite energy in this gas of electrons, there has to be a limit on the brightness temperature (which is determined by the density of electrons). This is a self-regulating process--if the brightness temperature goes too high, an infinite energy demand is set up, knocking it back down.

\subsection*{ The Sunyaev-Zeldovich Effect: Compton Y-Parameter}

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.

\end{document} <\latex>