BR of semileptonic B meson decay

[Safinaz]

Hi there,

In a reference as 1303.5877v1 [hep-ph ] the SM branching ratio of $B \to \tau \nu$ is given by:

$\frac{m_B G_F^2 m_\tau^2 \tau_B f^2_B } { 8 \pi } V_{ub}^2 ( 1 - \frac{m_\tau^2}{m_B^2} )^2$ . In the SM model the value of this BR $\sim 0.7 \times 10^{-4}$ .

But I don't understand how the BR is of order 10^-4, while the mean life time $\tau_B \sim 10^{-12} s$ and the Fermi constant $\sim 10^{-5} GeV ^{ -2}$ ?

 

[ChrisVer]

$10^{-12} ~s \sim \mathcal{O}(10^{11} ~GeV^{-1})$

*edit*

Well I got this from wolframalpha, but I think it's wrong...it's not always correct with its conversions...

eg. here:
http://www.saha.ac.in/theory/palashbaran.pal/conv.html
I find $10^{-12} ~s \approx 1.52 \times 10^{12} ~GeV$

I guess the right way is to set the natural units $\hbar= 6.58 \times 10^{-25} ~GeV~s=1$ so from that you can calculate by yourself the relation between energy and seconds.
You get:
$10^{-12} s = \frac{1}{6.58} 10^{13} ~GeV^{-1}= 1.52 \times 10^{12}~GeV^{-1}$

 

[ChrisVer]

The units then work fine, since the Branching ratio is immediately dimless...
$GeV^{-4}_{G_F} \times GeV_{m_E} \times GeV^{-1}_{\tau_B} \times GeV^2_{m_\tau} \times GeV^2_{f_B} =1$

and then you have $10^{-10}$ from $G_F^2$,
$10^{12}$ from $\tau_B$,
$10^{-5} \text{-} 10^{-6}$ from $|V_{ub}|^2$
so the orders of magnitude can work out. I don't know what values they used for the masses and the coupling constant f

 

[Safinaz]

Thanx ..