In class, we discussed Compton scattering $+$ q $$ $+$ q$,$ where a photon elastically hits a stationary charged particle q $($commonly electron$)$ and undergoes a wavelength shift due to energy loss in the interaction. Now, you will consider a different version of this, namely inverse Compton scattering. In this case, a relatively low$-$energy photon collides elastically with a high$-$energy charged particle, as a result of which yields a final photon more energetic than the initial one $($energy gain, or energy transfer$).$ Take c $=1$ below. a$)(25$ Points$)$ Consider a photon with energy $($E$\_$$)$ making a head$-$on collision $($i$.$e$.$ the incident angle $=$ $)$ with a proton with energy E$\_$p $>$ E$\_$$.$ Determine the scattering angle at which the energy of the final photon is maximum. b$)(10$ Points$)$ For a large energy proton, E$\_$p $>>$ m$\_$p$,$ show that the energy of the final photon $($for the maximal case you found in part a$)$ is given as E$\text{'}\_$ $=$ E$\_$p$/(1+($m$\_$p$)^2/4*$E$\_$p$*$E$\_$$).$ c$)(10$ Points$)$ Now consider ultra$-$high$-$energy cosmic protons with energy E$\_$p $=10^20$ eV$.$ Such rays are not expected to arrive at Earth if they are created outside our galaxy, since while traveling such distances $($larger than their mean free path$)$ they lose energy by interacting at least once with low$-$energy photons in the cosmic microwave background $($CMB$),$ which fills our universe. By using m$\_$p $10^9$ eV and CMB photon energy E $310^4$ eV with the result in part b$,$ estimate the energy loss ratio for the proton after a single interaction with a CMB photon.