# Radiation Lecture 17

<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} \subsection*{ Inverse Compton Scattering}

Recall the following rules for photons bouncing off of relativistic electrons: $$\begin{aligned}E_1&\sim\gamma^2E\\ max(E_1)&\sim\gamma m_ec^2\\ \end{aligned}$$ We'd like next to discuss the behavior of a single $e^-$ swimming through a sea of photons. We'd like to know how much power this $e^-$ is going to scatter by Compton-upscattering these photons. To order of magnitude, the power scattered by a single relativistic ($\gamma\gg1$) electron should depend on the cross-section for scattering, the speed of the electron ($c$), the \# density of photons having various energies, and the energy these photons take from scattering off of the electron: $$\begin{aligned}P&\sim\sigma_Tc\int{\eta(E)dE\,E_1(E)}\\ &\sim\sigma_Tc\int{\eta(E)dE\,\gamma^2E}\\ &\sim\gamma^2c\sigma_T U_{ph}\\ \end{aligned}$$ where $U_{ph}$ is the energy density of the photon field. This looks very similar to the power radiated by the synchrotron magnetic field, which is $P_{synch}\sim U_Bc\sigma_T\gamma^2$.\par This was an order of magnitude derivation, but we can do better--we just need to be more careful about how much the photons end up with, versus how much the electron actually gave to the photons (they had energy to begin with). We made the assumption that the initial energy of the photons was negligible because the electron was relativistic. However, what follows will be true for any $\gamma$.\par First, note that the energy bequeathed to the photon bath is given by $P_{net}=P_{scat}-P_{incident}$: $$\begin{aligned}P_{inc}&=c\sigma_T\int{\eta(E)E\,dE}\\ &=c\sigma_TU_{ph}\\ \end{aligned}$$ The power scattered {\it out} by the electron should be equal to the power radiated by the $e^-$ in its own rest frame, due to accelerations caused by $\ef^\prime$-fields of photons seen in its rest frame (recall that power is Lorentz-invariant quantity). In the $e^-$'s rest frame, the $\bfield$'s of the photons don't produce accelerations (there's no velocity), so all we need to consider are the Lorentz-transformed $\ef$'s of the photons: $$P^\prime={2\over3}{e^2(a^\prime)^2\over c^3} ={2\over3}{e^2\over c^3}\left({e\ef^\prime\over m_e}\right)^2 ={2\over3}{e^4\over m_e^2c^3}(\ef^\prime)^2$$ Recall that the Lorentz-transformed $\ef$'s look like: $$\begin{aligned}E_x^\prime&=E_x\\ E_y^\prime&=\gamma E_y-{\gamma v\over c}B_z\\ E_z^\prime&=\gamma E_z+{\gamma v\over c}B_y\\ \end{aligned}$$ for an electron moving in the $\^x$ direction. Substituting these values into our equation for $P^\prime$: $$\begin{aligned}\mean{P^\prime}&={2\over3}{e^4\over m_e^2c^3}\left(\mean{E_x^2}+ \gamma^2\mean{E_y^2}+\gamma^2\beta^2\mean{B_z^2}-\underbrace{2\gamma\beta \mean{E_yB_z}}_{=0}+\gamma^2\mean{E_z^2}+\gamma^2\beta^2\mean{B_y^2} +\underbrace{2\gamma^2\beta\mean{E_zB_y}}_{=0}\right)\\ &={2\over3}{e^4\mean{E_x^2}\over m_e^2c^3}\left(1+2\gamma^2+2\gamma^2\beta^2 \right)\\ &=\underbrace{{2\over3}{e^4\over m_e^2c^4}}_{=\sigma_T\over4\pi} {c\over3}\underbrace{3\mean{E_x^2}\over4\pi}_{=U_{ph}} 4\pi(1+2\gamma^2+2\gamma^2\beta^2)\\ &=\sigma_TcU_{ph}{1\over3}(2\gamma^2+2(\gamma^2-1)+1)\\ \end{aligned}$$ Recall that $\gamma^2\beta^2=\gamma^2-1$, so $\mean{P^\prime}=\sigma_T cU_{ph}{1\over3}(4\gamma^2-1)$, and our net power scattered is: $$\boxed{P_{net}={4\over3}\sigma_TU_{ph}c\beta^2\gamma^2}$$ This is an exact expression. Note that if $\gamma\gg1$ then the average photon energy of Compton-upscattered radiation is: $$\begin{aligned}\mean{h\nu}_1&={P\over{U_{ph}\over\mean{h\nu}}\sigma_Tc}\\ &=\mean{h\nu}{4\over3}\gamma^2\\ \end{aligned}$$ Another note: Compare our expression for the Compton-upscattering radiation to the power radiated by a single electron undergoing synchrotron radiation: $$P_{sync}=2\sigma_TcU_B\beta^2\gamma^2\mean{\sin^2\alpha}$$ The only difference (ignoring the $\sin^2\alpha$), is the exchange of $U_B$ for $U_{ph}$.\par

\subsection*{ 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 $\gamma\sim\left({\nu\over
\tn}\right)^\hf\propto\tn^{-\hf}$.

\end{document}
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