Правка разделов «Statement», «Solution», «Answer»

jzmicer правка от
Посмотреть решение История
правка #18501 предыдущая #18500 ← раньше
en/14.5.20.md +57 −3
@@ -1,11 +1,65 @@
### Statement
−$14.5.20.$ [Insert the problem statement]
+$14.5.20.$ Determine the minimum and maximum energies of the neutrino produced in the decay of a $\pi^0$-meson with energy $6\ \text{GeV}$:
+$$
+\pi^0\to\mu^+ + e^- + \nu
+$$
### Solution
+The neutrino is a massless particle, so for it $E_{\nu}=p_{\nu}c$.
−1
+The energy of the $\pi^0$-meson is much larger than its rest energy, so the total momentum of the system is nonzero. Then the invariant is:
+$$
+I = (m_{\pi^0}c)^2 = \frac{E^2}{c^2} - p_{\pi^0}^2 \quad\Rightarrow\quad p_{\pi^0} = \sqrt{\frac{E^2}{c^2} - (m_{\pi^0}c)^2} \tag{1}
+$$
+Clearly, the neutrino energy is maximum when the particle acquires the maximum momentum. This occurs when the neutrino in the laboratory frame receives momentum along the direction of motion of the $\pi^0$-meson, and the remaining particles move in the opposite direction with the same velocity. Note that the electron and antimuon can be considered as a single particle with effective rest mass $m_{\mu}+m_e$. You can prove this yourself; similar ideas are used in other problems in this section.
+$$
+p = \frac{E_{max}}{c} = p_{\mu e} + p_{\pi^0} \quad\Rightarrow\quad p_{\mu e}^2 = \left( \frac{E_{max}}{c} - p_{\pi^0} \right)^2 \tag{2}
+$$
+$$
+E = E_{max} + E_{\mu e} \quad\Rightarrow\quad E_{\mu e}^2 = (E - E_{max})^2 \tag{3}
+$$
+Now the invariant:
+$$
+(m_{\mu}+m_e)^2 c^2 = \frac{E_{\mu e}^2}{c^2} - p_{\mu e}^2 \tag{4}
+$$
+Substituting (2) and (3) into (4):
+$$
+(m_{\mu}+m_e)^2 c^2 = \frac{(E - E_{max})^2}{c^2} - \left( \frac{E_{max}}{c} - p_{\pi^0} \right)^2
+$$
+$$
+(m_{\mu}+m_e)^2 c^4 = E^2 - 2E_{max}(E - p_{\pi^0}c) - p_{\pi^0}^2 c^2 \tag{5}
+$$
+Using (1) in (5):
+$$
+(m_{\mu}+m_e)^2 c^4 = E^2 - 2E_{max}(E - p_{\pi^0}c) - E^2 + m_{\pi^0}^2 c^4
+$$
+$$
+2E_{max}\left( E - \sqrt{E^2 - m_{\pi^0}^2 c^4} \right) = m_{\pi^0}^2 c^4 - (m_{\mu}+m_e)^2 c^4
+$$
+$$
+2E_{max}\left( E - \sqrt{E^2 - m_{\pi^0}^2 c^4} \right) = m_{\pi^0}^2 c^4 - (m_{\mu}+m_e)^2 c^4
+$$
+$$
+E_{max} = \frac{m_{\pi^0}^2 c^4}{2} \frac{1 - \frac{(m_{\mu}+m_e)^2}{m_{\pi^0}^2}}{E - \sqrt{E^2 - m_{\pi^0}^2 c^4}}
+$$
+Multiply by $E + \sqrt{E^2 - m_{\pi^0}^2 c^4}$:
+$$
+E_{max} = \frac{m_{\pi^0}^2 c^4}{2} \frac{1 - \frac{(m_{\mu}+m_e)^2}{m_{\pi^0}^2}}{m_{\pi^0}^2 c^4} \left( E + \sqrt{E^2 - m_{\pi^0}^2 c^4} \right)
+$$
+$$
+E_{max} = \frac{E}{2} \left[ 1 - \frac{(m_{\mu}+m_e)^2}{m_{\pi^0}^2} \right] \left[ 1 + \sqrt{1 - \left( \frac{m_{\pi^0}c^2}{E} \right)^2} \right] \approx 2.3\ \text{GeV}
+$$
+The minimum energy obviously tends to zero: it is easy to show that the decay is possible without the neutrino's contribution according to energy and momentum conservation. And given that, according to modern concepts, this particle does have some rest mass, this energy can be strictly zero.
+
+Note: Savchenko has a mistake in his answer; he forgot to divide by 2.
+
#### Answer
−[Insert a concise answer or boxed result]
+$$
+E_{max}=\frac{E}{2}\left[1-\frac{(m_{\mu}+m_e)^2}{m_{\pi^0}^2}\right]\left[1+\sqrt{1-\left(\frac{m_{\pi^0}c^2}{E}\right)^2}\right]\approx 2.3\ \text{GeV}
+$$
+$$
+E_{min}=0
+$$