- #1
Tolya
- 22
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First, sorry for my English. I'm not very well in it... Please, try to understand :)
The problem.
We have reaction: \mu \rightarrow e + \nu + \tilde{\nu}
We know energy of myuon - E.
Question: Find the maximum and the minimum energy of electron.
My attept:
Conservation of energy: E = E_e+E_{\nu}+E_{\tilde{\nu}} (1)
Conservation of impulse: \vec{p} = \vec{p_e}+\vec{p_{\nu}}+\vec{p_{\tilde{\nu}}}} (2)
The mass of the rest of neutrino and antineutrino is 0. So, E_{\nu}=p_{\nu}c
E_{\tilde{\nu}}=p_{\tilde{\nu}}c and from the first equation:
E_e = E-c(p_{\nu}+p_{\tilde{\nu}})
Therefore, we must find the minumum and the maximum value of (p_{\nu}+p_{\tilde{\nu}}) Then, the minmum value of this expression gives us the maximum value of E_e and the maximum value gives the minimum of energy. Am I right in this statement?
From (2):
\vec{p_{\nu}}+\vec{p_{\tilde{\nu}}}} = \vec{p} - \vec{p_e}
Also we know that: p= \sqrt{\frac{E^2}{c^2}-m^2c^2}
and: p_e= \sqrt{\frac{{E_e}^2}{c^2}-{m_e}^2c^2}
How can I find the minumum and the maximum value of (p_{\nu}+p_{\tilde{\nu}}) with the help of all I wrote here? :)
I also have an assumption that we can find half of the answer simply in the following way:
E_e reaches its maximum when impulses of neutrino and antineutrino have opposite directions and equals in absolute. (It's easy to understand this fact because in this case impulses of neutrino and antineutrino compensate each other and the value p_e reaches its maximum, so does E_e). Then, almost easy:
p=\sqrt{\frac{E^2}{c^2}-m^2c^2}=p_e=\sqrt{\frac{{E_e}^2}{c^2}-{m_e}^2c^2}
And we have: E_e^{max}=\sqrt{E^2+c^4({m_e}^2-m^2)} Am I right? How can I find the minimum value of E_e?
Please, help.
The problem.
We have reaction: \mu \rightarrow e + \nu + \tilde{\nu}
We know energy of myuon - E.
Question: Find the maximum and the minimum energy of electron.
My attept:
Conservation of energy: E = E_e+E_{\nu}+E_{\tilde{\nu}} (1)
Conservation of impulse: \vec{p} = \vec{p_e}+\vec{p_{\nu}}+\vec{p_{\tilde{\nu}}}} (2)
The mass of the rest of neutrino and antineutrino is 0. So, E_{\nu}=p_{\nu}c
E_{\tilde{\nu}}=p_{\tilde{\nu}}c and from the first equation:
E_e = E-c(p_{\nu}+p_{\tilde{\nu}})
Therefore, we must find the minumum and the maximum value of (p_{\nu}+p_{\tilde{\nu}}) Then, the minmum value of this expression gives us the maximum value of E_e and the maximum value gives the minimum of energy. Am I right in this statement?
From (2):
\vec{p_{\nu}}+\vec{p_{\tilde{\nu}}}} = \vec{p} - \vec{p_e}
Also we know that: p= \sqrt{\frac{E^2}{c^2}-m^2c^2}
and: p_e= \sqrt{\frac{{E_e}^2}{c^2}-{m_e}^2c^2}
How can I find the minumum and the maximum value of (p_{\nu}+p_{\tilde{\nu}}) with the help of all I wrote here? :)
I also have an assumption that we can find half of the answer simply in the following way:
E_e reaches its maximum when impulses of neutrino and antineutrino have opposite directions and equals in absolute. (It's easy to understand this fact because in this case impulses of neutrino and antineutrino compensate each other and the value p_e reaches its maximum, so does E_e). Then, almost easy:
p=\sqrt{\frac{E^2}{c^2}-m^2c^2}=p_e=\sqrt{\frac{{E_e}^2}{c^2}-{m_e}^2c^2}
And we have: E_e^{max}=\sqrt{E^2+c^4({m_e}^2-m^2)} Am I right? How can I find the minimum value of E_e?
Please, help.
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