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Jun 6th, 2018
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  1. \documentclass[a4paper,11pt,twoside]{article}
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  38. \rhead{CHAPTER 14. COMPTON SCATTERING}
  39. \lhead{172}
  40. \cfoot{}
  41. \setcounter{section}{13}
  42.  
  43.  
  44.  
  45. \begin{document}
  46.     %Marcin Flis 17:30 27.05
  47.     \section{a}
  48.    
  49.     \begin{equation}
  50.     \label{14.21}
  51.     P_{-} = \langle h\nu \rangle \frac{d N}{dt} = c\sigma_{T} u
  52.     \tag{14.21}
  53.     \end{equation}
  54.    
  55.     \begin{equation}
  56.     \label{14.7}
  57.     P_{-} = \langle h\nu \rangle \frac{d N}{dt} = c\sigma_{T} u
  58.     \tag{14.7}
  59.     \end{equation}
  60.    
  61.     \begin{equation}
  62.     \label{14.19}
  63.     P_{-} = \langle h\nu \rangle \frac{d N}{dt} = c\sigma_{T} u
  64.     \tag{14.19}
  65.     \end{equation}
  66.    
  67.     \newpage
  68.    
  69.     Now Thomson scattering is independent of frequency of photon, so it follows that the rate at which
  70.     energy is scattered out of radiation field is given by multiplying $dN/dt$ by the mean energy per
  71.     photon, to give
  72.     \begin{equation}
  73.     \label{14.24}
  74.     P_{-} = \langle h\nu \rangle \frac{d N}{dt} = c\sigma_{T} u
  75.     \tag{14.24}
  76.     \end{equation}
  77.    
  78.     Note that this is precisely the rate at which energy is scattered for a stationary electron. This is
  79.     something of a coincidence since we can see from (\ref{14.21}) that the radiation field by moving
  80.     electron do \textit{not} have an isotrophic distribution in the lab-frame, rather they have a $( 1-
  81.     \beta \cos \theta)$ distribution, so the photons propagating in the direction opposite to the electron
  82.     are more likely to be scattered.
  83.    
  84.     Combining $P_{+} $ from (\ref{14.19}) and $ P_{-}$ form (\ref{14.24}) gives the net
  85.     \textit{inverse Compton power} for 1 electron of $ P = P_+ - P_- $ or
  86.    
  87.     \begin{equation}
  88.     \label{14.25}
  89.     P = \frac{4}{3} \beta^2 \gamma^2 c \sigma_{T} u
  90.     \tag{14.25}
  91.     \end{equation}
  92.    
  93.     and the total energy transfer rate per unit volume is given by mulitiplying this by the elctron density,
  94.     or more generally by the distrobution function $ n(\textbf{r},E)$ and intergrating over energy.
  95.    
  96.     Equation (\ref{14.25}) is remarkably simple, and also ramerkably similar to the synchrotron power
  97.     and bremsstrahlung power, for reasons already disscussed.
  98.    
  99.     Interstingly, for low velocities, the Compton power is quadratic in the velocity. There is no first
  100.     order-effect, since a scatterings may increase or decrease the photon energy.
  101.    
  102.     \setcounter{subsection}{3}
  103.     \subsection{Compton vs Inverse Compton Scattering}
  104.     Equation (\ref{14.25}) is supposedly valid for all electron eneregies, and is clealry always positive.
  105.     However, this does not make sense. For cold electrons, Compton scattering result in a loss of energy
  106.     for the elsctrons vis the recoil, which was ignored in deriving (\ref{14.25}).
  107.    
  108.     For the low eneregy electrons (with $ v/c = \beta \ll 1$), the radiation in the electron frame is very
  109.     nearly isotropic ($\delta \nu / \nu \sim \beta \ll 1$, so consequently the variation of intensity $\delta
  110.     I/I \sim \beta \ll 1 $ and is also small, so we can incorpotate the effect of recoil by simply
  111.     subtracting the meran photon energy loss givern by (\ref{14.7}).
  112.    
  113.     For $\upsilon \ll c $ mean rate of energy transfer is given by $ dE/ dt \backsimeq (4/3)c \sigma_{T}
  114.     u \upsilon^2 / c^2 $ while the rate of scatterings is $dN/dt = c \sigma_{T} n = c \sigma_{T} u /
  115.     \langle \epsilon \rangle $ so the mean photon eneregy gain per collision (neglecting recoil) is $
  116.     \langle \Delta \epsilon \rangle / \langle \epsilon \rangle = (4/3)(v/c)^2 $, and if $v \ll c $ that is
  117.     approximately equal to the mean \textit{frractional} energy gain $ \langle \Delta \epsilon /
  118.     \epsilon \rangle = (4/3)(v/c)^2 $ For a thermal distribution of electrons this becomes $\langle \Delta
  119.     \epsilon / \epsilon \rangle = 4kT/ mc^2 $. The mean fractional eneregy loss due to recoil if from
  120.     (\ref{14.7}) $ \Delta \epsilon / \epsilon = \epsilon/ Mc^2 $ so combining these gives
  121.    
  122.     \begin{equation}
  123.     \label{14.26}
  124.     \left\langle \frac{\Delta \epsilon}{\epsilon} \right\rangle = \frac{4kT - h\omega}{m c^2}
  125.     \tag{14.26}
  126.     \end{equation}
  127.    
  128.     It $\epsilon > 4kT $ then there is net transfer of energy to the electrons and \textit{vice versa}.
  129.    
  130.    
  131.     \subsection{ The Compton $y$- Parameter}
  132.    
  133.     The \textit{ Compton y-parameter } is defined as
  134.    
  135.     \begin{equation*}
  136.     \label{14}
  137.     y \equiv \left\langle \frac{\Delta \epsilon}{\epsilon} \right\rangle \times \langle \text{number of
  138.         scatterings} \rangle
  139.     \end{equation*}
  140.    
  141.     \begin{itemize}
  142.         \item In a system with $y$ much less (greater) then unity spactrum will be little (strongly)
  143.         affected by scattering(s).
  144.         \item In computing $y$ it is usual to either use the non relativistic expression (\ref{14.26})
  145.         or the higlhy relativistic limit $ \left \langle \Delta \epsilon / \epsilon \right\rangle \simeq 4
  146.         \gamma^2 /3 $.
  147.         \item The mean number of scatterings is given by $ \text{max}(\tau,\tau^2)$.
  148.     \end{itemize}
  149.    
  150. \end{document}
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