Synchrotron Self-Interactions

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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} $${C\jn\over\pmag^{3\over4}\nu_1^{-\hf}}\sim\pmag$$ $${B^2\over8\pi}=\pmag\sim(C\jn\nu_1^\hf)^{4\over7}$$

\subsection*{ Synchrotron Self-Absorption}

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.