What Causes the Observed Excess in the Semileptonic Decay Rate of B Mesons?

  • Thread starter Thread starter Safinaz
  • Start date Start date
  • Tags Tags
    Decay Mesons
Safinaz
Messages
255
Reaction score
8
Hi there,

Since some time I started to study the semilptonic decay of B meson : ## B \to D \tau \nu ## that there is an excess in the value R (D) = ## \frac{ Br ( B \to D \tau \nu) } {Br (B \to D l \nu) } ## in BABAR experiment than the SM expectations- See for example arXiv:1302.7031v5- Now I have two questions about the numirecal results of such decay, if anyone has an experience about ..

* Since Br = the partial decay width/ the total decay width, won't R(D) = ## \frac{d \Gamma ( B \to D \tau \nu) }{ d q^2 d \cos \theta_l } ÷ \frac{d \Gamma ( B \to D l \nu) }{ d q^2 d \cos \theta_l} ## ?

According to [1302.7031v5] notation. Where q^2 is the centre of mass energy squared and ## \cos \theta_l ## is the angle of the lepton relative to the B rest frame.

* If we used ## d q^2 = 2 q dq ##, are the integration limits for ## \cos \theta ## : 0 < ## \cos\theta ## < ## \pi ## and for q: 0 < q < ## \sqrt{10} ## .. I think if we set all NP couplings to zero in Equ. 2.2 [1302.7031v5] we should have the SM value of R(D) ..

Thanx.
 
Last edited:
Physics news on Phys.org
Oh, an interesting deviation.

2012 they studied it without considering the angular distributions, 2013 they took that into account.
You can always consider the partial decay width for the decay channel (integrated over its own phase-space), then the ratio is given by the ratios of branching fractions.
What is an angle relative to a rest frame?

Theory paper link. If you set all NP contributions to zero you get the SM, sure.
 
Hi, the attached papers are so useful .. but I think the angle ## \theta_l ## is just the angle of the emitted ## \tau ## or lepton regarding to B meson .. it can be ## d_0 ## in Equ. 8 [ 1303.0571], so in all cases we should integrate over , whether it set in a formula tells about the angular distribution or not.

Also notice that the NP 2HDM contribution to ## \Gamma ## (B-> D tau nu) in [ 1303.0571] is considered by multiplying the SM amplitude by a factor ~ tan theta/ mH+ , While in other references as 1302.7031v5, the NP considered starting by adding new couplings to the SM effective Lagrangian.
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Replies
3
Views
2K
Replies
13
Views
5K
Replies
13
Views
3K
Replies
49
Views
12K
Replies
5
Views
4K
Replies
6
Views
2K
Back
Top