# Muon momentum

1. Nov 24, 2007

### indigojoker

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

in the decay process, $$\pi ^{+} -> \mu^{+} + \nu_{\mu}$$

show that for a neutrino of finite (but small) mass, compared with the case of the massless neutrino, the muon momentum would be reduced by the fraction:

$$\frac{p'}{p}= - \frac{m_{\nu}^2 (m_{\pi}^2+m_{\mu}^2)}{(m_{\pi}^2-m_{\mu}^2)^2}$$

Calculation of the muon's momentum with massless neutrino:

$$p_{\mu}=p_{\pi}-p_{\nu}$$
$$p_{\mu}^2=p_{\pi}^2+p_{\nu}^2-2m_{\pi}E_{\nu}$$
We know:
$$p_{\mu}=m_{\mu}^2c^2$$
$$p_{\pi}=m_{\pi}^2c^2$$
$$p_{\nu}=0$$

$$m_{\nu}^2c^2=m_{\pi}^2c^2-2m_{\pi}E_{\nu}$$

But: $$E_{\nu}=|p_{\nu}|c=|p_{\mu}|c$$

$$2 m_{\pi} |p_{\mu}|=(m_{\pi}^2-m_{\mu}^2)c$$
$$|p_{\mu}|=\frac{m_{\pi}^2-m_{\mu}^2}{2m_{\pi}}c$$

Using the same method, but with $$p_{\nu}=m_{\nu}^2c^2$$

we get:
$$2 m_{\pi} |p_{\nu}|=(m_{\pi}^2-m_{\mu}^2-m_{\nu}^2)c$$
$$|p'_{\mu}|=\frac{m_{\pi}^2-m_{\mu}^2-m_{\nu}^2}{2m_{\pi}}c$$

The fraction I get is:

$$\frac{p'}{p}=\frac{m_{\pi}^2-m_{\mu}^2-m_{\nu}^2}{m_{\pi}^2-m_{\mu}^2}$$

If i multiply by $$\frac{m_{\pi}^2-m_{\mu}^2}{m_{\pi}^2-m_{\mu}^2}$$ I do not get the above result. Any idea where i went wrong?

Last edited: Nov 24, 2007