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### Statement
−$14.5.22.$ [Insert the problem statement]
+$14.5.22.$ What is the maximum energy that photons with energy $\mathcal{E} = 10 \text{ eV}$ can acquire when scattered on a counter-propagating electron beam with energy $\mathcal{E}_e = 10^{10} \text{ eV}$?
### Solution
+The problem can be solved simply by writing down the conservation of momentum and energy, but I want to show another method. One can notice that if we go to a reference frame where the electrons in the beam are at rest, we observe the [Compton effect](https://en.wikipedia.org/wiki/Compton_scattering). The change in energy upon scattering of a photon in this case can be obtained from the well-known formula
−$14.5.22.$ What is the maximum energy that photons with energy $\mathcal{E} = 10 \text{ eV}$ can acquire when scattered on a counter-propagating electron beam with energy $\mathcal{E}_e = 10^{10} \text{ eV}$?
+$$
+\frac{1}{\nu'}-\frac{1}{\nu}=\frac{h}{m_e c^2}(1-\cos\alpha)
+$$
+where $\alpha$ is the scattering angle of the photon. This formula is derived in problem [$14.5.23$](https://savchenkosolutions.com/en/14.5.23).
+As a result, we need to carefully transform to such a system, calculate the energy, and then transform back to the laboratory frame.
+
+For convenience, we switch to a system of units where $c = 1$. This simplifies the notation, and it is convenient that the photon momentum equals its energy. We will denote energies in the laboratory frame (LF) by $\mathcal{E}$, and in the electron rest frame (RF) by $E$.
+
+Thus, the gamma factor for the electron in the LF is obviously
+\gamma=(1-v^2)^{-\frac{1}{2}}=\frac{ \mathcal{E}_e }{m_e}, \ \to \ v=\sqrt{1-\left(\frac{m_e}{ \mathcal{E}_e }\right)^2}=\frac{\sqrt{\mathcal{E}_e^2-m_e^2}}{\mathcal{E}_e}
+$$
+This is the speed at which the RF moves relative to the LF. We find the photon energy in the RF using the relativistic Doppler effect:
+$$
+E=\mathcal{E}\gamma(1+v)
+$$
+
+Using the Compton formula, we obtain the energy change upon collision:
+$$
+E'=\frac{Em_e}{E(1-\cos\alpha)+m_e}
+$$
+
+
+The maximum energy in the LF is achieved when the photon is scattered backward in the RF ($\alpha = \pi$), because upon inverse transformation to the LF, a photon scattered backward in the RF receives the maximum Doppler shift forward (in the direction of the electron's motion):
+$$
+E'=\frac{Em_e}{2E+m_e}
+$$
+Inverse transformation (taking into account that the photon's direction has reversed):
+$$
+\mathcal{E}'=E'\gamma(1+v)
+$$
+$$
+\mathcal{E}'_{max} =\frac{Em_e}{2E+m_e}\gamma(1+v)=\frac{\mathcal{E}\gamma^2(1+v)^2m_e}{2\mathcal{E}\gamma(1+v)+m_e}
+$$
+Multiply numerator and denominator by $(1 - v)$:
+$$
+\mathcal{E}'_{max} =\frac{\mathcal{E}m_e(1+v)}{2\mathcal{E}\gamma(1-v^2)+m_e(1-v)}
+$$
+$$
+\mathcal{E}'_{max} =\frac{\mathcal{E}m_e(1+v)\gamma}{2\mathcal{E}+m_e\gamma(1-v)}
+$$
+$$
+\mathcal{E}'_{max} =\mathcal{E}\frac{\mathcal{E}_e(1+v)}{2\mathcal{E}+\mathcal{E}_e(1-v)}
+$$
+Finally substitute the velocity:
+$$
+\mathcal{E}'_{max} =\mathcal{E}\frac{\mathcal{E}_e+\sqrt{\mathcal{E}_e^2-m_e^2}}{2\mathcal{E}+\mathcal{E}_e-\sqrt{\mathcal{E}_e^2-m_e^2}}
+$$
+
+Now return to standard units:
+
+$$
+\mathcal{E}'_{max} =\mathcal{E}\frac{\mathcal{E}_e+\sqrt{\mathcal{E}_e^2-(m_e c^2)^2}}{2\mathcal{E}+\mathcal{E}_e-\sqrt{\mathcal{E}_e^2-(m_e c^2)^2}}
+$$
+Don't forget the numerical value (Savchenko forgot...):
+$$
+\mathcal{E}_{max}\approx 6\cdot10^9 \ эВ
+$$
+Note: Savchenko apparently has a small typo in the answer.
+
+
#### Answer
−[Insert a concise answer or boxed result]
+$$
+\boxed{
+\mathcal{E}_{max}=\mathcal{E}\frac{\mathcal{E}_e+\sqrt{\mathcal{E}_e^2+(m_e c^2)^2}}{2\mathcal{E}+\mathcal{E}_e-\sqrt{\mathcal{E}_e^2-(m_e c^2)^2}}\approx6\cdot10^9 \ эВ
+}
+$$