| @@ -1,11 +1,75 @@ |
| | | ### 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 |
| | + | } |
| | + | $$ |
