# Compton Scattering

### Reference Material

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\section*{ Compton Scattering}

Compton scattering is the inelastic scattering of a photon off of an electron. This is the quantum mechanical or high-energy extension to Thomson scattering.

If the photon {\it loses} energy, its wavelength will increase and this is called {\bf Compton Scattering}. If the electron has sufficient initial kinetic energy, the photon the photon can {\it gain} energy, this is called {\bf Inverse Compton Scattering}, or Compton Up-Scattering. An alteration of the photon spectrum (either up-scattering or down-scattering) due to interactions with electrons is called comptonization\par

Some examples of applications of Compton scattering are: \begin{itemize} \item Compton exchange keeps electrons in thermal equilibrium with photons at redshifts $z\ge10^3$. \item The spectra of AGN and xray binaries are altered by Compton Scattering (e.g. radio emission to optical wavelengths). \item CMB photons get upscattered by galaxy cluster plasma. This is called the Sunyaev-Zeldovich effect. \item Inverse Compton scattering has been proposed as a likely emission mechanism for gamma ray bursts. \end{itemize}

The Compton effect was first observed in the 1920's and Arthur Holly Comton was awarded the 1927 Nobel prize for the discovery. The effect is often regarded as some of the first concrete experimental evidence for quantum mechanics. \par

\subsection*{Mathematical Derivation of Compton Scattering}

To find the final energy of a scattered photon and the Compton shift (the change in the photon wavelength), we use the conservation of energy and momentum.

In the standard derivation of Compton scattering, the electron is assumed to be free and at rest. This is a good approximation considering the photon energies for which this process is significant are much larger than relevant electron binding energies.

A photon with initial energy $E_{\gamma_i} = h \nu_i$ travelling in the $\hat{x}$ direction scatters of an electron at rest ($E_{e_{i}} = m_ec^2$). After the scatter, the photon has energy $E_{\gamma_f} = h \nu_f$ and is travelling at an angle $\theta$ relative to the original $\hat{x}$ direction. The electron has energy given by $E_{e_f} = \sqrt{ p_e^2 c^2 + m_e^2 c^4}$ and has scattered at a different angle. See figure below for a depiction of the relevant variables.

The conservation of energy tells us \begin{aligned} E_{\gamma_i} + E_{e_i} &= E_{\gamma_f} + E_{e_f} \\ E_{\gamma_i} + m_ec^2 &= \sqrt{p_e^2 c^2 + m_e^2 c^4} + E_{\gamma_f} \end{aligned} Rearranging and squaring both sides gives $$(E_{\gamma_i} - E_{\gamma_f} + m_ec^2)^2 = p_e^2 c^2 + m_e^2 c^4. \label{eq:inter}$$ We will use the conservation of momentum to write the $p_e^2 c^2$ factor in terms of the photon energies.

The conservation of momentum tells us $$\vec{p_{\gamma_i}} = \vec{p_{\gamma_f}} + \vec{p_{e_f}}$$ $$(\vec{p_{\gamma_i}} - \vec{p_{\gamma_f}} )^2 = \vec{p_{e_f}}^2$$ $$p_{\gamma_i}^2 + p_{\gamma_f}^2 - 2 p_{\gamma_i} p_{\gamma_f} \cos \theta = p_{e_f}^2 \label{eq:2}$$ where $\theta$ is the angle between the initial photon direction $\hat{x}$ and the final photon direction.

Multiplying Eqn.~\ref{eq:2} by $c^2$ and using the relation $E_\gamma = p_\gamma c$, we can rewrite it as $$E_{\gamma_i}^2 + E_{\gamma_f}^2 - 2 E_{\gamma_i}E_{\gamma_f}\cos\theta = p_e^2 c^2.$$ Now, we can plug this into Eqn.~\ref{eq:inter} above and expand the squared brackets on the left side at the same time: $${E_{\gamma_i}^2} + {E_{\gamma_f}^2} + {m_e^2c^4} + {2} E_{\gamma_i}m_ec^2 - {2}E_{\gamma_f} m_ec^2 - {2}E_{\gamma_i}E_{\gamma_f} = {E_{\gamma_i}^2} + {E_{\gamma_f}^2} - {2}E_{\gamma_i}E_{\gamma_f}\cos\theta + {m_e^2c^4}.$$ Crossing off similar terms on both sides gives us $$E_{\gamma_i} m_e c^2 - E_{\gamma_f} m_e c^2 - E_{\gamma_i}E_{\gamma_f} = E_{\gamma_I}E_{\gamma_f} \cos \theta.$$

Solving for the final photon energy gives $$E_{\gamma_f} = \frac{E_{\gamma_i}}{1+\frac{E_{\gamma_i}}{m_ec^2}(1-\cos\theta)}, \label{eq:comptonenergy}$$ and the change in the photon wavelength, referred to as the Compton shift, is found to be $$\frac{1}{E_{\gamma_f}} - \frac{1}{E_{\gamma_i}} = \frac{1}{m_ec^2}(1-\cos\theta)$$ $$\lambda_f - \lambda_i = \frac{h}{m_e c}(1-\cos\theta) \label{eq:comptonwavelength}$$

The prefactor $\lambda = \frac{h}{m_e c}$ is called the Compton wavelength. $\lambda_c = 0.02$~\r{A} gives the order of change in wavelength during a Compton scatter interaction; therefore, Compton scattering is only relavent for high energy (low wavelength) photons.

Comments: \begin{itemize} \item $\lambda_1-\lambda>0$: the shift here is tiny. The maximum possible value is $\lambda_1-\lambda= 2h m_ec=0.04$\r{A}. The momentum tends to be shared between the electron and photon, but no so much the energy. \item Remember this is {\it scattering}, not absorption. Photon \# is conserved. \item If $\lambda_i \gg \lambda_c$ (or $h\nu \ll m_e c^2$), the scattering can be approximated as elastic (Thomson scattering) and $\Delta E_\gamma = 0$. \item The derivation for the cross section of Compton scattering, which is given by the Klein-Nishina formula, is outside of the scope of [R \& L]. An important thing to know is that the scattering angle is anisotropic and depends on energy. In the limit that $h_\nu \gg m_e c^2$, then $\sigma \sim \sigma_T\left(\frac{m_e c^2}{h \nu}\right)$, and this additional term is called the Klein-Nishina correction. \end{itemize}

\subsection*{Inverse Compton Scattering}

When the electron has significant kinetic energy as compared to the energy of the photon, then energy can be transferred from the electron to the photon, resulting in the scattered photon having a higher frequency.

To find the final energy of the photon, we must perform a couple Lorentz transformations. The standard Compton equations (3 \& 4 above) are valid from the electron's rest frame. The angles and frequencies of the photon as seen by the relativistic electron are different than what is measured in the lab frame.

What we've done so far was for a stationary electron. For a moving electron, we need to consider the dependence on the angle at which the photon is coming in with respect to the direction of velocity ($\theta$). To make this situation similar to the one we just considered, we need to be in the frame of the electron. In this frame, the photons has a new energy as a result of time dilation: $$E^\prime=\gamma E(1-{\frac{v}{ c}}\cos\theta)$$ The $\frac{v}{c}$ term is just the classic Doppler shift of the photon. Now suppose that the photon (which entered at angle $\theta^\prime$ in the electron frame) rebounds at an angle $\theta_1^\prime$. To relate this to our previous derivation, we want to find $\phi$. So note that $\theta_1^\prime-\phi^\prime=\theta^\prime$. Thus, in the electron's frame: $$E_1^\prime={E^\prime \frac{1}{1+{E^\prime \frac{1}{ m_ec^2}(1-\cos\phi^\prime)}}}$$ Transforming this back into the lab frame: $$E_1=E_1^\prime\gamma(1+\frac{v}{ c}\cos\theta_1^\prime)$$ This generally follows Rybicki \& Lightman. However, it might help to know that in R\&L, 7.7b follows from 7.8a, which follows from 7.7a.\par Comments: \begin{itemize} \item If $E^\prime\ll m_ec^2$, then: \begin{aligned}E_1&\approx E^\prime\gamma(1+\frac{v}{c}\cos\theta_1^\prime)\\ &\approx\gamma^2E(1+\frac{v}{c}\cos\theta_1^\prime)(1-\frac{v}{c}\cos\theta)\\ \end{aligned} In a typical collision, $\theta\sim\theta_1^\prime\sim \frac{\pi}{2}$, so $E_1\sim\gamma^2E$. \item If $E^\prime\gg m_ec^2$, then: \begin{aligned}E_1^\prime&={E^\prime \frac{1}{1+{E^\prime \frac{1}{ m_ec^2}}} (1-\cos\phi^\prime)}\approx m_ec^2\\ E_1&=E_1^\prime\gamma(1+\frac{v}{c}\cos\theta_1^\prime)\\ &\approx m_ec^2\gamma(1+\frac{v}{c}\cos\theta_1^\prime)\approx\gamma m_ec^2\\ \end{aligned} This final term defines the maximum rebound of the photon. \end{itemize}

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