# Compton scattering by cosmic rays

As an example of Compton scattering, we can consider the case where a photon from the cosmic microwave background scatters off a high-energy proton travelling as a cosmic ray. In the Sun’s reference frame, we observe a photon of energy ${h\nu=2\times10^{-4}}$ eV scattering off a proton with energy ${10^{9}m_{P}}$, where ${m_{P}}$ is the rest mass of the proton, which is around ${10^{9}}$ eV. We can find the highest energy attainable by the photon after scattering off the proton.

In order to use the Compton scattering formula, we need to work in the rest frame of the proton. In the Sun’s frame, the energy of the proton is ${10^{9}m_{P}}$ so ${\gamma}$ for transforming to the rest frame of the proton is ${\gamma=10^{9}}$. This corresponds to a velocity of

 $\displaystyle v$ $\displaystyle =$ $\displaystyle \sqrt{1-\frac{1}{\gamma^{2}}}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 1-\frac{1}{2}\times10^{-18} \ \ \ \ \ (2)$

where the approximation uses the first two terms of a Taylor expansion.

We can now transform to the initial rest frame of the proton, but we need to specify the relative velocities of the photon and proton. Since we want to find the highest possible energy for the scattered photon, we have to choose the scattering angle that will give this energy. The greatest energy transfer to the photon will occur when the two particles scatter directly back on themselves (that is, the collision is ‘head-on’, so the photon exactly reverses its original direction). We can therefore take the photon to be heading in the ${+x}$ direction and the proton in the ${-x}$ direction. Transforming to the proton’s rest frame means transforming to a frame moving in the ${-x}$ direction at speed ${v}$. Since this frame is heading in the opposite direction to the photon, the photon will appear blue-shifted towards a higher frequency, by the Doppler factor (assuming ${v>0}$) ${\sqrt{(1+v)/(1-v)}\approx2\times10^{9}}$.

The energy of the photon in the proton’s rest frame is therefore ${2\times10^{9}\times2\times10^{-4}eV=4\times10^{5}\; eV}$.

A back-scatter occurs when ${\theta=\pi}$, so using the Compton formula,

 $\displaystyle \frac{1}{h\nu_{f}}$ $\displaystyle =$ $\displaystyle \frac{1}{h\nu_{i}}+\frac{1-\cos\theta}{m_{P}}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.25\times10^{-5}+2\times10^{-9}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 0.25\times10^{-5}\; eV^{-1}\ \ \ \ \ (5)$ $\displaystyle h\nu_{f}$ $\displaystyle =$ $\displaystyle 4\times10^{5}\; eV \ \ \ \ \ (6)$

where ${\nu_{f}}$ and ${\nu_{i}}$ are the final and initial frequencies, respectively.

To transform back to the original frame (the initial proton rest frame), we use the same value for ${v}$, but since the photon has reversed direction after scattering, its frequency in the original frame is again blue-shifted relative to the proton’s rest frame. That is, the scattered frequency back in the Sun’s frame is

 $\displaystyle h\nu_{f(sun)}$ $\displaystyle =$ $\displaystyle 4\times10^{5}\times2\times10^{9}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 8\times10^{14}\; eV \ \ \ \ \ (8)$

Planck’s constant in units of ${eV-s}$ is ${h=4\times10^{-15}\; eV-s}$, so we get

$\displaystyle \nu_{f(sun)}=2\times10^{29}s^{-1} \ \ \ \ \ (9)$

which is (very) far into the gamma-ray range.