# Kinematics of two body decay

1. Apr 19, 2015

### Safinaz

Hi there,

In a two body decay process like $B \to l \nu$, there is a factor $| \bar{l} (1-\gamma_5)\nu|^2$ in the matrix element amplitude, in which equals

$| \bar{l} (1-\gamma_5)\nu|^2 = (/\!\!\! p_l+m_l) (1-\gamma_5) /\!\!\! p_\nu(1+\gamma_5) = 2(/\!\!\! p_l+m_l) /\!\!\! p_\nu(1+\gamma_5) = 8 p_l. p_\nu ~$(1) , after taking the trace.

To evaluate (1) I set the process kinematics as

$p_l = ( E_l,\bf{p_l})$ and $p_\nu= ( E_\nu,- E_\nu)$, where

$-E_\nu = -\bf{p_\nu} \equiv - \bf{p_l}, ~ \bf{p_l}^2= E_l^2 - m_l^2, ~ E_l =E_\nu = m_B/2$, then

$p_l. p_\nu = E_l E_\nu +\bf{p_l}^2 = 2E_l^2 - m_l^2 = m_B^2/ 2 - m_l^2 = m_B^2/ 2 ( 1 - \frac{2m_l^2}{m_B^2})$, so I got a factor
$( 1 - \frac{2m_l^2}{m_B^2})$, while in references as [hep-ph/0306037v2] equ. 5, (1) gave a factor $( 1 - \frac{m_l^2}{m_B^2})$ instead,

Bests.

2. Apr 19, 2015

### vanhees71

Let's start with the kinematics. In the rest frame of the decaying B meson you have
$$p_l=(E_l,\vec{p}), \quad p_{\nu}=(E_{\nu},-\vec{p}).$$
This means
$$s=(p_l+p_{\nu})^2=m_{B}^2$$
but
$$s=m_B^2=m_l^2+m_{\nu}^2+2 p_l \cdot p_{\nu},$$
which gives
$$2 p_l \cdot p_{\nu}=m_B^2-m_l^2-m_{\nu}^2 \simeq m_B^2-m_l^2=m_B^2 \left (1-\frac{m_l^2}{m_B^2} \right).$$
Your mistake is setting $E_l$ and $E_{\nu}$ equal. However the magnitudes of the momenta are equal, not the energies.

To get the cm. energies we use that
$$s=m_l^2+m_{\nu}^2+2(E_l E_{\nu}+P^2)=m_l^2+m_{\nu}^2+2 (E_l \sqrt{s}-m_1^2) \; \Rightarrow \; E_l=\frac{m_B^2+m_l^2-m_{\nu}^2}{2m_B}.$$
In the 2nd step we've used $\sqrt{s}=E_1+E_2$. Exchanging the labels $l$ and $\nu$ in the above calculation gives
$$E_{\nu}=\frac{m_B^2+m_{\nu}^2-m_{l}^2}{2m_B} \neq E_l.$$

Last edited: Apr 19, 2015
3. Apr 19, 2015

Many thanks.