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Inclusive/Exclusive measurement of B-mesons

  1. Jul 10, 2015 #1

    ChrisVer

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    What is the differences between inclusive and exclusive B-meson measurements?
    I think the inclusive is that in the case eg of a [itex]B\bar{B}[/itex] mesons creation, we use the one ##B## meson as a tag (and measure all its products) while we don't measure the mesonic decay of the other ##B## but only the lepton originating from its decay... What is the exclusive then?
    And why don't we measure everything?

    Thanks
     
  2. jcsd
  3. Jul 10, 2015 #2

    mfb

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    Staff: Mentor

    "Inclusive B meson" => there was at least one B-meson, the decay channel and possible other particles do not matter. The measurement method will depend on the detector.
    "Exclusive" => needs more specification what got measured. Probably a specific decay channel.

    All those analyses should describe what they measure, you can look it up there (which is much more precise than the general concept anyway).
     
  4. Jul 10, 2015 #3

    ChrisVer

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    For example in the [itex]V_{ub}[/itex] measurement, one can use both inclusive and exclusive measurements in order to determine this element's value.

    In the inclusive measurement, one looks at [itex]B \rightarrow X_u l \nu[/itex], and distinguishes the signal by using the [itex]q^2=(p_l + p_\nu)^2[/itex] momentum transfer of the [itex]W[/itex] and the [itex]m_X[/itex] reconstructed invariant mass of the final hadronic system. It seems in this case we don't care about the final hadron system (Except for its invariant mass).
    On the other hand in the exclusive measurement, one can look at [itex]B \rightarrow \pi l \nu[/itex] and measures the [itex]q^2[/itex] (I think [itex]m_\pi[/itex] is known so we don't care about it?) and the rate.

    One strange thing is that for the last one looks at [itex]\frac{d\Gamma (B\rightarrow \pi l \nu)}{dq^2}[/itex] while for the first at [itex]\Gamma(B \rightarrow X_u l \nu)[/itex] alone. I think I saw the same thing for other measurements too..
    The first depends on the momentum of the pion:
    \begin{equation}
    \frac{d\Gamma}{dq^2} (B \rightarrow \pi \nu l) = \frac{G_F^2}{24 \pi^3} p_\pi^3 |V_{ub}|^2 |f_+(q^2)|^2
    \end{equation}

    While the second does not (but it has dependencies on different factors coming from EW and QCD):

    \begin{equation}

    \Gamma (B \rightarrow X_u l \nu) = \frac{G_F^2}{192 \pi^3} m_b^5 |V_{ub}|^2 A_{ew} A_{pert} A_{non-pert}

    \end{equation}
     
  5. Jul 10, 2015 #4
    I always thought in flavour physics, the inclusive / exclusive measurements refer to b-> q transitions.

    For example, studying the B-> Kmumu is exclusive. (Only 1 particular final state from a b->s transition).

    While studying the B-> all final States involving an s quark would be inclusive for b->s.

    Theoretically, computing such b->s transitions inclusively is often more simple.

    Unlike experimental physics, the term has a different meaning.
     
  6. Jul 10, 2015 #5
    Then let me add, the exclusive decay has differential information about that particular final state. So the prediction of this observable is final state dependent, which is why it appears for exclusive and not inclusive.

    I guess inclusive predictions are more for branching fractions etc.
     
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