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Kinematics of two body decay

  1. Apr 19, 2015 #1
    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,
    So what's wrong I made?

    Bests.
     
  2. jcsd
  3. Apr 19, 2015 #2

    vanhees71

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    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
  4. Apr 19, 2015 #3
    Many thanks.
     
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