# Relativistic Kinematics Problem

1. May 15, 2007

### Flux = Rad

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.

1. The problem statement, all variables and given/known data

A pion travelling at speed $$v$$ decays into a muon and a neutrino,
$$\pi^- \rightarrow \mu^- + \bar{\nu}_\mu$$
If the neutrino emerges at 90 degrees to the original pion direction, at what angle does the muon come off?

2. Relevant equations

Relativistic formulae for energy and momentum, energy and momentum conservation.

3. The attempt at a solution

Note: in the following I've set $$c = 1$$. 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 travelling in the $$\hat{x}$$ direction, and the neutrino goes off in the $$\hat{y}$$ direction.

The four vectors for the particles are:
$$P_\pi = \left( \begin{array}{cc} E_\pi / c \\ p_\pi \\ 0 \\ 0 \end{array} \right) \qquad P_\nu = \left( \begin{array}{cc} p_\nu \\ 0 \\ p_\nu \\ 0 \end{array} \right) \qquad P_\mu = \left( \begin{array}{cc} E_\mu / c \\ p_x \\ p_y \\ 0 \end{array} \right)$$

4 momentum conservation states that

$$P_\pi = P_\nu + P_\mu$$

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:

$$p_y = \frac{2 E_\mu p_\nu - m_\pi ^2 + m_\mu ^2}{2 p_\nu}$$

If I rearrange my initial statement of energy/momentum conservation to

$$P_\nu = P_\pi - P_\mu$$

the result of squaring this is

$$p_x = \frac{2 E_\pi E_\mu - m_\pi ^2 m_\mu ^2}{2 p_\pi}$$

This allows me to obtain an expression for the required angle, namely

$$\tan \theta = \frac{p_y}{p_x} = \frac{p_\pi (2 E_\mu p_\nu - m_\pi ^2 + m_\mu ^2)}{p_\nu (2 E_\pi E_\mu - m_\pi ^2 - m_\mu ^2)}$$

In order to get rid of the $$p_\nu$$ terms, I take my conservation master equation and rearrange it to the last possible arrangement, and square:

$$P_\mu = P_\pi - P_\nu \quad \Rightarrow \quad P_\nu = \frac{m_\pi ^2 - m_\mu ^2}{2 E_\pi}$$

Substituting this into my rather messy equation for $$\tan \theta$$, I end up with:

$$\tan \theta = 2 \frac{p_\pi (E_\mu - E_\pi)}{2 E_\pi E_\mu - m_\pi ^2 m_\mu ^2}$$

Clearly this isn't satisfactory since we don't know what $$E_\mu$$ is. I tried using the energy-mass-momentum relation to remedy this but I only managed to get

$$\tan \theta = 2 \frac{p_\pi(E_\mu - E_\pi)}{p_\mu ^2 - p_\pi ^2}$$

which isn't of much help either, since we now have unknown $$p_\mu$$ terms floating about.

Last edited: May 15, 2007
2. May 15, 2007

### variation

Hello,

I try this way:
$$P_\pi = \left( \begin{array}{cc} E_\pi / c \\ p_\pi \\ 0 \\ 0 \end{array} \right) \qquad P_\nu = \left( \begin{array}{cc} p_\nu \\ 0 \\ p_\nu \\ 0 \end{array} \right) \qquad P_\mu = \left( \begin{array}{cc} E_\mu / c \\ p_x \\ p_y \\ 0 \end{array} \right)$$
Actually,conservation of momentum give $$p_\pi=p_x$$ and $$p_\nu=p_y$$
Therefore,
$$\tan\theta=\frac{p_y}{p_x}=\frac{p_\nu}{p_\pi}$$
where
$$p_\pi=\gamma m_\pi v$$
and
$$p_\nu = \frac{m_\pi ^2 - m_\mu ^2}{2 E_\pi}$$ (you have got this relation)
Simplify further
$$\tan\theta=\frac{p_y}{p_x}=\frac{p_\nu}{p_\pi}=\frac{m_\pi ^2 - m_\mu ^2}{2 E_\pi}\frac{1}{\gamma m_\pi v}=\frac{m_\pi ^2 - m_\mu ^2}{2 \gamma m_\pi}\frac{1}{\gamma m_\pi v}=\frac{1}{2\gamma^2 v}\left(\frac{m_\pi ^2 - m_\mu ^2}{m^2_\pi}\right)$$

Best regards

Last edited: May 15, 2007
3. May 15, 2007

### Dick

Without following all of the details (sorry, it's getting late), but you haven't used at all the fact that you were given the initial velocity of the pion is 'v'. This gives you a relation between E_pi and p_pi in terms of v. That's one less variable.

Last edited: May 15, 2007
4. May 16, 2007

### Flux = Rad

Thanks for the replies, I was making things way too complicated for myself. I'm not sure if the expression I got was correct but variation's method is far more elegant and uhm... logical. Hope that'll teach me to keep an eye on the bigger picture with simpler problems.

Thanks again.