Правка разделов «Statement», «Solution», «Answer»
en/14.5.17.md
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| ### Statement | |||
| − | $14.5.17.$ [Insert the problem statement] | ||
| + | $14.5.17$. At what minimum kinetic energy of a positron does its collision with a stationary electron cause the production of a proton-antiproton pair: $e^+ + e^- \rightarrow p + \bar{p}$? How many times is this energy larger than the minimum kinetic energy of a positron that produces a proton-antiproton pair in a head-on collision with an electron? | ||
| ### Solution | |||
| − | 1 | ||
| − | #### Answer | ||
| + | The author suggests two ways to produce a proton-antiproton pair. We interpret the conditions as in the second case the electron and positron move toward each other with equal speeds. | ||
| − | |||
| + | We also remember that the minimum energy corresponds to the final particles being at rest relative to each other. Then the invariant in the first case is: | ||
| + | |||
| + | $$ | ||
| + | I_1 = \frac{(E_{e} + m_e c^2)^2}{c^2} - p_e^2 = \frac{(2m_p c^2)^2}{c^2} = 4m_p^2 c^2 \tag{1} | ||
| + | $$ | ||
| + | |||
| + | where $E_{e} = m_e c^2+E_{k1}$ and $p_e = \gamma m_e v = \frac{\beta}{c} E_e $. One could set up complicated equations in terms of $\beta$, but it is easier to write another invariant, this time for the positron: | ||
| + | |||
| + | $$ | ||
| + | m_e^2 c^2 = \frac{E_e^2}{c^2} - p_e^2 \tag{2} | ||
| + | $$ | ||
| + | |||
| + | Substituting (2) into (1): | ||
| + | |||
| + | $$ | ||
| + | \frac{(E_e + m_e c^2)^2}{c^2} - \frac{E_e^2}{c^2} = 4m_p^2 c^2 - m_e^2 c^2 | ||
| + | $$ | ||
| + | |||
| + | $$ | ||
| + | 2E_e m_e c^2 = 4m_p^2 c^4 - 2m_e^2 c^4 | ||
| + | $$ | ||
| + | |||
| + | $$ | ||
| + | E_e = c^2 \frac{2m_p^2 - m_e^2}{m_e} | ||
| + | $$ | ||
| + | |||
| + | $$ | ||
| + | E_{k1} = E_e - m_e c^2 = 2c^2 \frac{m_p^2 - m_e^2}{m_e} \tag{3} | ||
| + | $$ | ||
| + | |||
| + | Now we consider the second case similarly, noting that the total momentum of the system is zero: | ||
| + | |||
| + | $$ | ||
| + | I_2 = \frac{(2E_e)^2}{c^2} = \frac{(2m_p c^2)^2}{c^2} = 4m_p^2 c^2 \tag{4} | ||
| + | $$ | ||
| + | |||
| + | $$ | ||
| + | E_e = m_p c^2 \tag{5} | ||
| + | $$ | ||
| + | |||
| + | $$ | ||
| + | E_{k2} = c^2 (m_p - m_e) | ||
| + | $$ | ||
| + | |||
| + | $$ | ||
| + | N = \frac{E_{k1}}{E_{k2}} = 2 \frac{m_p^2 - m_e^2}{m_e (m_p - m_e)} = 2 \left( \frac{m_p}{m_e} + 1 \right) \approx 3.7 \cdot 10^3 | ||
| + | $$ | ||
| + | It is clear that almost all the kinetic energy in the first case goes not to the creation of the pair, but to its acceleration. | ||
| + | |||
| + | Note: Savchenko has a typo in the second part of the answer, but it hardly affects the numerical result. | ||
| + | #### Answer | ||
| + | $$ | ||
| + | \boxed{E_{k1}=2c^2\frac{m_p^2-m_e^2}{m_e}, \qquad N=2\left(\frac{m_p}{m_e}+1\right)=3.7\cdot 10^3} | ||
| + | $$ | ||
| @@ -1,11 +1,63 @@ | |||
| ### Statement | ### Statement | ||
| $14.5.17.$ [Insert the problem statement] | $14.5.17$. At what minimum kinetic energy of a positron does its collision with a stationary electron cause the production of a proton-antiproton pair: $e^+ + e^- \rightarrow p + \bar{p}$? How many times is this energy larger than the minimum kinetic energy of a positron that produces a proton-antiproton pair in a head-on collision with an electron? | ||
| ### Solution | ### Solution | ||
| 1 | |||
| #### Answer | The author suggests two ways to produce a proton-antiproton pair. We interpret the conditions as in the second case the electron and positron move toward each other with equal speeds. | ||
| We also remember that the minimum energy corresponds to the final particles being at rest relative to each other. Then the invariant in the first case is: | |||
| $$ | |||
| I_1 = \frac{(E_{e} + m_e c^2)^2}{c^2} - p_e^2 = \frac{(2m_p c^2)^2}{c^2} = 4m_p^2 c^2 \tag{1} | |||
| $$ | |||
| where $E_{e} = m_e c^2+E_{k1}$ and $p_e = \gamma m_e v = \frac{\beta}{c} E_e $. One could set up complicated equations in terms of $\beta$, but it is easier to write another invariant, this time for the positron: | |||
| $$ | |||
| m_e^2 c^2 = \frac{E_e^2}{c^2} - p_e^2 \tag{2} | |||
| $$ | |||
| Substituting (2) into (1): | |||
| $$ | |||
| \frac{(E_e + m_e c^2)^2}{c^2} - \frac{E_e^2}{c^2} = 4m_p^2 c^2 - m_e^2 c^2 | |||
| $$ | |||
| $$ | |||
| 2E_e m_e c^2 = 4m_p^2 c^4 - 2m_e^2 c^4 | |||
| $$ | |||
| $$ | |||
| E_e = c^2 \frac{2m_p^2 - m_e^2}{m_e} | |||
| $$ | |||
| $$ | |||
| E_{k1} = E_e - m_e c^2 = 2c^2 \frac{m_p^2 - m_e^2}{m_e} \tag{3} | |||
| $$ | |||
| Now we consider the second case similarly, noting that the total momentum of the system is zero: | |||
| $$ | |||
| I_2 = \frac{(2E_e)^2}{c^2} = \frac{(2m_p c^2)^2}{c^2} = 4m_p^2 c^2 \tag{4} | |||
| $$ | |||
| $$ | |||
| E_e = m_p c^2 \tag{5} | |||
| $$ | |||
| $$ | |||
| E_{k2} = c^2 (m_p - m_e) | |||
| $$ | |||
| $$ | |||
| N = \frac{E_{k1}}{E_{k2}} = 2 \frac{m_p^2 - m_e^2}{m_e (m_p - m_e)} = 2 \left( \frac{m_p}{m_e} + 1 \right) \approx 3.7 \cdot 10^3 | |||
| $$ | |||
| It is clear that almost all the kinetic energy in the first case goes not to the creation of the pair, but to its acceleration. | |||
| Note: Savchenko has a typo in the second part of the answer, but it hardly affects the numerical result. | |||
| #### Answer | |||
| $$ | |||
| \boxed{E_{k1}=2c^2\frac{m_p^2-m_e^2}{m_e}, \qquad N=2\left(\frac{m_p}{m_e}+1\right)=3.7\cdot 10^3} | |||
| $$ | |||