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I've done a countless number of these problems before, so I was quite annoyed when I was unable to do this one. It's from Griffiths' Intro to Elementary Particles.
A pion traveling at speed [tex]v[/tex] decays into a muon and a neutrino,
[tex] \pi^- \rightarrow \mu^- + \bar{\nu}_\mu [/tex]
If the neutrino emerges at 90 degrees to the original pion direction, at what angle does the muon come off?
Relativistic formulae for energy and momentum, energy and momentum conservation.
Note: in the following I've set [tex]c = 1[/tex]. If that bothers anyone too much, I could do it again keeping the c's in, but it's rather tedious.
I've set it up so that the pion was traveling in the [tex] \hat{x} [/tex] direction, and the neutrino goes off in the [tex] \hat{y} [/tex] direction.
The four vectors for the particles are:
[tex] P_\pi = \left( \begin{array}{cc} E_\pi / c \\ p_\pi \\ 0 \\ 0 \end{array} \right) \qquad P_\nu = \left( \begin{array}{cc} p_\nu \\ 0 \\ p_\nu \\ 0 \end{array} \right) \qquad P_\mu = \left( \begin{array}{cc} E_\mu / c \\ p_x \\ p_y \\ 0 \end{array} \right) [/tex]
4 momentum conservation states that
[tex] P_\pi = P_\nu + P_\mu [/tex]
Dotting both sides of this equation with itself, and noting that
a) the square of a 4 momentum is the rest mass of the particle
b) the neutrino is effectively massless
I get:
[tex] p_y = \frac{2 E_\mu p_\nu - m_\pi ^2 + m_\mu ^2}{2 p_\nu} [/tex]
If I rearrange my initial statement of energy/momentum conservation to
[tex] P_\nu = P_\pi - P_\mu [/tex]
the result of squaring this is
[tex] p_x = \frac{2 E_\pi E_\mu - m_\pi ^2 m_\mu ^2}{2 p_\pi} [/tex]
This allows me to obtain an expression for the required angle, namely
[tex] \tan \theta = \frac{p_y}{p_x} = \frac{p_\pi (2 E_\mu p_\nu - m_\pi ^2 + m_\mu ^2)}{p_\nu (2 E_\pi E_\mu - m_\pi ^2 - m_\mu ^2)} [/tex]
In order to get rid of the [tex] p_\nu [/tex] terms, I take my conservation master equation and rearrange it to the last possible arrangement, and square:
[tex] P_\mu = P_\pi - P_\nu \quad \Rightarrow \quad P_\nu = \frac{m_\pi ^2 - m_\mu ^2}{2 E_\pi} [/tex]
Substituting this into my rather messy equation for [tex] \tan \theta [/tex], I end up with:
[tex] \tan \theta = 2 \frac{p_\pi (E_\mu - E_\pi)}{2 E_\pi E_\mu - m_\pi ^2 m_\mu ^2} [/tex]
Clearly this isn't satisfactory since we don't know what [tex] E_\mu [/tex] is. I tried using the energy-mass-momentum relation to remedy this but I only managed to get
[tex] \tan \theta = 2 \frac{p_\pi(E_\mu - E_\pi)}{p_\mu ^2 - p_\pi ^2} [/tex]
which isn't of much help either, since we now have unknown [tex] p_\mu [/tex] terms floating about.
Homework Statement
A pion traveling at speed [tex]v[/tex] decays into a muon and a neutrino,
[tex] \pi^- \rightarrow \mu^- + \bar{\nu}_\mu [/tex]
If the neutrino emerges at 90 degrees to the original pion direction, at what angle does the muon come off?
Homework Equations
Relativistic formulae for energy and momentum, energy and momentum conservation.
The Attempt at a Solution
Note: in the following I've set [tex]c = 1[/tex]. If that bothers anyone too much, I could do it again keeping the c's in, but it's rather tedious.
I've set it up so that the pion was traveling in the [tex] \hat{x} [/tex] direction, and the neutrino goes off in the [tex] \hat{y} [/tex] direction.
The four vectors for the particles are:
[tex] P_\pi = \left( \begin{array}{cc} E_\pi / c \\ p_\pi \\ 0 \\ 0 \end{array} \right) \qquad P_\nu = \left( \begin{array}{cc} p_\nu \\ 0 \\ p_\nu \\ 0 \end{array} \right) \qquad P_\mu = \left( \begin{array}{cc} E_\mu / c \\ p_x \\ p_y \\ 0 \end{array} \right) [/tex]
4 momentum conservation states that
[tex] P_\pi = P_\nu + P_\mu [/tex]
Dotting both sides of this equation with itself, and noting that
a) the square of a 4 momentum is the rest mass of the particle
b) the neutrino is effectively massless
I get:
[tex] p_y = \frac{2 E_\mu p_\nu - m_\pi ^2 + m_\mu ^2}{2 p_\nu} [/tex]
If I rearrange my initial statement of energy/momentum conservation to
[tex] P_\nu = P_\pi - P_\mu [/tex]
the result of squaring this is
[tex] p_x = \frac{2 E_\pi E_\mu - m_\pi ^2 m_\mu ^2}{2 p_\pi} [/tex]
This allows me to obtain an expression for the required angle, namely
[tex] \tan \theta = \frac{p_y}{p_x} = \frac{p_\pi (2 E_\mu p_\nu - m_\pi ^2 + m_\mu ^2)}{p_\nu (2 E_\pi E_\mu - m_\pi ^2 - m_\mu ^2)} [/tex]
In order to get rid of the [tex] p_\nu [/tex] terms, I take my conservation master equation and rearrange it to the last possible arrangement, and square:
[tex] P_\mu = P_\pi - P_\nu \quad \Rightarrow \quad P_\nu = \frac{m_\pi ^2 - m_\mu ^2}{2 E_\pi} [/tex]
Substituting this into my rather messy equation for [tex] \tan \theta [/tex], I end up with:
[tex] \tan \theta = 2 \frac{p_\pi (E_\mu - E_\pi)}{2 E_\pi E_\mu - m_\pi ^2 m_\mu ^2} [/tex]
Clearly this isn't satisfactory since we don't know what [tex] E_\mu [/tex] is. I tried using the energy-mass-momentum relation to remedy this but I only managed to get
[tex] \tan \theta = 2 \frac{p_\pi(E_\mu - E_\pi)}{p_\mu ^2 - p_\pi ^2} [/tex]
which isn't of much help either, since we now have unknown [tex] p_\mu [/tex] terms floating about.
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