Synchrotron Self-Compton

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\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} \title{ Synchrotron Self-Compton (SSC)}

Electrons undergoing synchrotron radiation create a photon bath which other electrons will then interact with via inverse Compton scattering. Recall that for original (unprocessed) synchrotron radiation, that $F_\nu$, between some minimum and maximum frequency cut-off, goes as $K\nu^\alpha$, and that the number of photons per $\gamma$ is ${dN\over d\gamma}=N_0\gamma^s$, where $\alpha={1+s\over2}$. These frequency cut-offs were set by $\gamma_{min}^2 \nu_{cyc}$ and $\gamma_{max}^2\nu_{cyc}$. After this radiation is processed by SSC, approximately every photon is upscattered to a new energy ${4\over3}\gamma^2\nu$. We are assuming that the relationship between an incoming photon frequency and it's final frequency are related via a delta function. Thus: \def\tnTemplate:\tilde\nu $$F_{\nu,SSC}(\nu)=\tau\int_{\tn}{K\tn^\alpha d\tn\delta\left(\tn- {\nu\over\gamma^2}\right)\int_\gamma{N_0\gamma^sd\gamma}}$$ Keep in mind that $N_0$ is normalized to so the integral comes out to 1 (it just accounts for the ``shape of the energy distribution function). $\tau$ is what contains the actual \# density of $e^-$'s. It is the fraction scattered, and is generally $\ll1$. $\nu\sim\tn\gamma^2$.\par For a fixed $\nu\sim\tn\gamma^2$, we find that $\gamma\sim\left({\nu\over \tn}\right)^\hf\propto\tn^{-\hf}$.

\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)} $$\begin{aligned}F_{\nu,SSC}&=\tau K\gamin^{2\alpha}\numin^\alpha N_0\ln\gllogul\\ &=\tau K\numin^\alpha\gamin^{2\alpha}\gamin^{-1-s}\ln\gllogul\\ &=\tau K \numin^\alpha\ln\gllogul\\ \end{aligned}$$

\end{document} <\latex>