- #1
Safinaz
- 259
- 8
Hi,
1. Homework Statement
In the calculation of the matrix element amplitude of ## B \to l \nu ##, I got a factor ## \bar{l} (1-\gamma_5)\nu ## as in [hep-ph/0306037v2]. For ##|M|^2## I made :
##| \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.
[/B]
My question about the value of (1 )which depends on the kinematics of the process.I got a factor
##\frac{m_B^2}{2} ( 1 - \frac{2m_l^2}{m_B^2}) ##, while in [hep-ph/0306037v2] equ. 5, (1) gave a factor ## m_B^2( 1 - \frac{m_l^2}{m_B^2}) ## instead ..
[/B]
I defined
## 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 =\frac{m_B^2}{2} ( 1 - \frac{2m_l^2}{m_B^2}) ##!
Bests.
1. Homework Statement
In the calculation of the matrix element amplitude of ## B \to l \nu ##, I got a factor ## \bar{l} (1-\gamma_5)\nu ## as in [hep-ph/0306037v2]. For ##|M|^2## I made :
##| \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.
Homework Equations
[/B]
My question about the value of (1 )which depends on the kinematics of the process.I got a factor
##\frac{m_B^2}{2} ( 1 - \frac{2m_l^2}{m_B^2}) ##, while in [hep-ph/0306037v2] equ. 5, (1) gave a factor ## m_B^2( 1 - \frac{m_l^2}{m_B^2}) ## instead ..
The Attempt at a Solution
[/B]
I defined
## 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 =\frac{m_B^2}{2} ( 1 - \frac{2m_l^2}{m_B^2}) ##!
Bests.