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- \documentclass[a4paper,11pt,twoside]{article}
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- \rhead{CHAPTER 14. COMPTON SCATTERING}
- \lhead{172}
- \cfoot{}
- \setcounter{section}{13}
- \begin{document}
- %Marcin Flis 17:30 27.05
- \section{a}
- \begin{equation}
- \label{14.21}
- P_{-} = \langle h\nu \rangle \frac{d N}{dt} = c\sigma_{T} u
- \tag{14.21}
- \end{equation}
- \begin{equation}
- \label{14.7}
- P_{-} = \langle h\nu \rangle \frac{d N}{dt} = c\sigma_{T} u
- \tag{14.7}
- \end{equation}
- \begin{equation}
- \label{14.19}
- P_{-} = \langle h\nu \rangle \frac{d N}{dt} = c\sigma_{T} u
- \tag{14.19}
- \end{equation}
- \newpage
- Now Thomson scattering is independent of frequency of photon, so it follows that the rate at which
- energy is scattered out of radiation field is given by multiplying $dN/dt$ by the mean energy per
- photon, to give
- \begin{equation}
- \label{14.24}
- P_{-} = \langle h\nu \rangle \frac{d N}{dt} = c\sigma_{T} u
- \tag{14.24}
- \end{equation}
- Note that this is precisely the rate at which energy is scattered for a stationary electron. This is
- something of a coincidence since we can see from (\ref{14.21}) that the radiation field by moving
- electron do \textit{not} have an isotrophic distribution in the lab-frame, rather they have a $( 1-
- \beta \cos \theta)$ distribution, so the photons propagating in the direction opposite to the electron
- are more likely to be scattered.
- Combining $P_{+} $ from (\ref{14.19}) and $ P_{-}$ form (\ref{14.24}) gives the net
- \textit{inverse Compton power} for 1 electron of $ P = P_+ - P_- $ or
- \begin{equation}
- \label{14.25}
- P = \frac{4}{3} \beta^2 \gamma^2 c \sigma_{T} u
- \tag{14.25}
- \end{equation}
- and the total energy transfer rate per unit volume is given by mulitiplying this by the elctron density,
- or more generally by the distrobution function $ n(\textbf{r},E)$ and intergrating over energy.
- Equation (\ref{14.25}) is remarkably simple, and also ramerkably similar to the synchrotron power
- and bremsstrahlung power, for reasons already disscussed.
- Interstingly, for low velocities, the Compton power is quadratic in the velocity. There is no first
- order-effect, since a scatterings may increase or decrease the photon energy.
- \setcounter{subsection}{3}
- \subsection{Compton vs Inverse Compton Scattering}
- Equation (\ref{14.25}) is supposedly valid for all electron eneregies, and is clealry always positive.
- However, this does not make sense. For cold electrons, Compton scattering result in a loss of energy
- for the elsctrons vis the recoil, which was ignored in deriving (\ref{14.25}).
- For the low eneregy electrons (with $ v/c = \beta \ll 1$), the radiation in the electron frame is very
- nearly isotropic ($\delta \nu / \nu \sim \beta \ll 1$, so consequently the variation of intensity $\delta
- I/I \sim \beta \ll 1 $ and is also small, so we can incorpotate the effect of recoil by simply
- subtracting the meran photon energy loss givern by (\ref{14.7}).
- For $\upsilon \ll c $ mean rate of energy transfer is given by $ dE/ dt \backsimeq (4/3)c \sigma_{T}
- u \upsilon^2 / c^2 $ while the rate of scatterings is $dN/dt = c \sigma_{T} n = c \sigma_{T} u /
- \langle \epsilon \rangle $ so the mean photon eneregy gain per collision (neglecting recoil) is $
- \langle \Delta \epsilon \rangle / \langle \epsilon \rangle = (4/3)(v/c)^2 $, and if $v \ll c $ that is
- approximately equal to the mean \textit{frractional} energy gain $ \langle \Delta \epsilon /
- \epsilon \rangle = (4/3)(v/c)^2 $ For a thermal distribution of electrons this becomes $\langle \Delta
- \epsilon / \epsilon \rangle = 4kT/ mc^2 $. The mean fractional eneregy loss due to recoil if from
- (\ref{14.7}) $ \Delta \epsilon / \epsilon = \epsilon/ Mc^2 $ so combining these gives
- \begin{equation}
- \label{14.26}
- \left\langle \frac{\Delta \epsilon}{\epsilon} \right\rangle = \frac{4kT - h\omega}{m c^2}
- \tag{14.26}
- \end{equation}
- It $\epsilon > 4kT $ then there is net transfer of energy to the electrons and \textit{vice versa}.
- \subsection{ The Compton $y$- Parameter}
- The \textit{ Compton y-parameter } is defined as
- \begin{equation*}
- \label{14}
- y \equiv \left\langle \frac{\Delta \epsilon}{\epsilon} \right\rangle \times \langle \text{number of
- scatterings} \rangle
- \end{equation*}
- \begin{itemize}
- \item In a system with $y$ much less (greater) then unity spactrum will be little (strongly)
- affected by scattering(s).
- \item In computing $y$ it is usual to either use the non relativistic expression (\ref{14.26})
- or the higlhy relativistic limit $ \left \langle \Delta \epsilon / \epsilon \right\rangle \simeq 4
- \gamma^2 /3 $.
- \item The mean number of scatterings is given by $ \text{max}(\tau,\tau^2)$.
- \end{itemize}
- \end{document}
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