Finding Maximum and Minimum Energy of Electron in Particle Disintegration

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In summary, the conversation revolves around a physics problem involving the conservation of energy and impulse in a reaction involving a muon, electron, neutrino, and antineutrino. The question is to find the maximum and minimum energy of the electron and various equations and assumptions are discussed to solve the problem. The conversation also briefly touches on the appropriate forum for posting this problem.
  • #1
Tolya
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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: [tex] \mu \rightarrow e + \nu + \tilde{\nu}[/tex]
We know energy of myuon - E.
Question: Find the maximum and the minimum energy of electron.

My attept:

Conservation of energy: [tex] E = E_e+E_{\nu}+E_{\tilde{\nu}}[/tex] (1)
Conservation of impulse: [tex] \vec{p} = \vec{p_e}+\vec{p_{\nu}}+\vec{p_{\tilde{\nu}}}}[/tex] (2)

The mass of the rest of neutrino and antineutrino is 0. So, [tex]E_{\nu}=p_{\nu}c[/tex]
[tex]E_{\tilde{\nu}}=p_{\tilde{\nu}}c[/tex] and from the first equation:

[tex] E_e = E-c(p_{\nu}+p_{\tilde{\nu}})[/tex]

Therefore, we must find the minumum and the maximum value of [tex](p_{\nu}+p_{\tilde{\nu}})[/tex] Then, the minmum value of this expression gives us the maximum value of [tex]E_e[/tex] and the maximum value gives the minimum of energy. Am I right in this statement?

From (2):
[tex] \vec{p_{\nu}}+\vec{p_{\tilde{\nu}}}} = \vec{p} - \vec{p_e} [/tex]
Also we know that: [tex]p= \sqrt{\frac{E^2}{c^2}-m^2c^2}[/tex]
and: [tex]p_e= \sqrt{\frac{{E_e}^2}{c^2}-{m_e}^2c^2}[/tex]

How can I find the minumum and the maximum value of [tex](p_{\nu}+p_{\tilde{\nu}})[/tex] 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:
[tex]E_e[/tex] 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 [tex]p_e[/tex] reaches its maximum, so does [tex]E_e[/tex]). Then, almost easy:
[tex]p=\sqrt{\frac{E^2}{c^2}-m^2c^2}=p_e=\sqrt{\frac{{E_e}^2}{c^2}-{m_e}^2c^2}[/tex]
And we have: [tex]E_e^{max}=\sqrt{E^2+c^4({m_e}^2-m^2)}[/tex] Am I right? How can I find the minimum value of [tex]E_e[/tex]?

Please, help.
 
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  • #2
No issue with the language, except that I am curious about what country considers beta or weak decay to be "Introductory Physics"
 
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  • #3
sweden do =)

And also, there are guys posting mass falling to the Earth without corilos and centrifugal terms in the advanced forum. Very hard to get everything right here..
 
  • #4
I'm very sorry, what forum corresponds to this theme?
 
  • #5
Tolya said:
I'm very sorry, what forum corresponds to this theme?

Well, it's not a problem really, but this topic is appropriate for the Advanced Physics HW forum. This is upper level material as opposed to introductory physiscs.
 
  • #6
Please, delete this theme. I posted this problem in "Advenced Physics".
 

1. What is the "disintegration of a particle"?

The disintegration of a particle refers to the process of a particle breaking down into smaller components or particles.

2. What causes a particle to disintegrate?

Particles can disintegrate through various means such as decay, collision, or interaction with other particles or energy.

3. How is the disintegration of a particle studied?

Scientists use various methods such as particle accelerators and detectors to study the disintegration of particles and understand the underlying processes.

4. What are the implications of particle disintegration?

The disintegration of particles is important in fields such as nuclear physics, astrophysics, and particle physics, as it helps us understand the fundamental building blocks of matter and the laws of the universe.

5. Can particle disintegration be controlled or harnessed?

While scientists are able to study and observe the process of particle disintegration, it is currently not possible to control or harness it for practical applications.

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