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Upsilon Meson Decays

  1. May 11, 2015 #1
    1. The problem statement, all variables and given/known data

    (a)Draw feynman diagrams of upsilon meson. Why is decay to ##q\bar q## states suppressed? Explain why width for 4s is much wider.
    (b) How do B mesons decay? Why is no other type of interaction possible? Draw feynman diagrams.
    (c) Find the distance travelled by 1s and 4s.
    (d) Why is asymmetry useful here? Find ##\beta##. Find ##E_+, E_-, \beta##.
    2009_B4_Q4.png

    2. Relevant equations


    3. The attempt at a solution

    Part(a)

    Feynman diagrams are given by
    2009_B4_Q4_2.png
    Decay to other ##q \bar q## is cabbibo suppressed by ##\sin \theta_c##.
    I'm not sure why the width at 4s is particularly wide.

    Part(b)
    Weak interactions. Only weak interactions are possible as the flavour is not conserved. Feynman diagrams are shown below:
    2009_B4_Q4_3.png

    Part(c)
    How can I work out ##c\tau## if I don't even have their lifetimes?
     
  2. jcsd
  3. May 14, 2015 #2
  4. May 14, 2015 #3

    mfb

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    Huh? Where did my post go.

    Why should it?
    There is no decay to ##q \bar q## present in your diagrams.

    Same comment as usual: check the particle masses to find possible decays. Which interactions are responsible for them?

    Right (but the W sign is odd, one B+ should be B- and the D- label are missing).

    You have their decay widths, that is nearly the same.
     
  5. May 15, 2015 #4
    True. Sorry I was thinking of inter-generation mixing in weak interactions. I think it is suppressed because strong interactions are preferred over weak ##(Z^0)## and EM ##(\gamma)## interactions. If that's the case, then wouldn't EM/weak production of lepton/anti-lepton pairs be suppressed as well?

    I think something happened between ##3s## and ##4s## state, something extra must have been produced between ##10.355GeV## and ##10.580 GeV##. I think it's the production of ##B^{+}B^{-}## pair of bosons.

    That's right. Thanks for pointing that out.

    Since ##\Delta E \sim \Gamma## and ##\Delta E \Delta \tau \sim \hbar##,their lifetimes are ##\tau \sim \frac{\hbar}{\Gamma}##. The values are ##\tau_{1s} = 1.3 \times 10^{-20} s## and ##\tau_{4s} = 2.7 \times 10^{-23}s##. The distance travelled before decaying is given by ##c\tau_{1s} = 3.8 \times 10^{-12}m## and ##c\tau_{4s} = 8.1 \times 10^{-15} m ##. The hierarchy is B mesons travel furthest, followed by 1s and then 4s. Not sure why this is the case.
     
  6. May 15, 2015 #5

    mfb

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    What is suppressed relative to what now?

    And ##B^0 \bar B^0##, right.

    Think of the available interactions for decays.
     
  7. May 15, 2015 #6
    Suppressed relative to strong interactions.

    1s and 4s decays via strong interaction while the B mesons that only decay via weak interactions (as flavour is not conserved) and that corresponds to longer lifetimes. Stronger the interaction, the shorter the lifetime.
     
  8. May 15, 2015 #7

    mfb

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    How does the 1s decay via strong interaction look like? That is exactly the qqbar decay that is missing here so far. And it is suppressed in the same way the charmonium you had a while ago rarely decays via the strong interaction.

    Right.
     
  9. May 15, 2015 #8
    image.png

    OZI suppression. That explains a factor of 1000 out, which is typically the case for OZI suppression.
     
  10. May 15, 2015 #9
    Part (d)
    Beam power asymmetry is useful as it avoids the centre of mass frame so new particles that are produced will move relative to the lab frame. This allows us to work out the production point of the decay.

    Energy of 4s particle is given by ##E = E_{-}(1+\delta)##. It's gamma factor is given by ##\gamma = \frac{E_{-}(1+\delta)}{M_{4s}}##. It's speed in beta is given by ##\beta = \left(1-\frac{1}{\gamma^2} \right)^{\frac{1}{2}}##. Distance it travels in lab frame is given by ##L = \left(c\beta\right) \left( \gamma \tau \right)## as we observe proper time to be dilated by a factor of ##\gamma##.

    Given that ##E=10.58## and ##\delta=0.1##, we find that ##E_{-} = 9.62~GeV##, ##E_{+}=0.96~GeV##, ##\gamma = 1.1##, ##\beta = 0.42## and ##L = 3.74 \times 10^{15} ~m##.
     
  11. May 15, 2015 #10

    mfb

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    Do you think 3.74*1015 m can be a reasonable answer?

    Also, your value for E looks wrong.
     
  12. May 15, 2015 #11
    Sorry I meant ##3.74 \times 10^{-15}m##.
     
  13. May 16, 2015 #12

    mfb

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    For a B+, this is way too short.
     
  14. May 16, 2015 #13
    I'm not sure what went wrong. Consider two beams smashing together, with one moving at energy equal to mass of particle to be produced, the another moving towards it at 10% of that energy. Total energy is ##1.1 \times M_{4s}##. The gamma factor of particle produced is correspondingly ##\gamma =1.1##. That gives a speed of ##0.46c##. Distance travelled is dilated time multiplied by speed, ##v (\gamma \tau)##. Using the lifetime of 4s as ##3 \times 10^{-23}~s##, I get distance travelled as ##3 \times 10^{-15}~m##.
     
  15. May 16, 2015 #14

    mfb

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    No, as the particle does not get produced at all. The center-of-mass energy is not sufficient. That is the wrong energy I mentioned.

    For the flight distance of a B+ you should use the B+ lifetime, not the Y(4s) lifetime.
     
  16. May 16, 2015 #15
    How is the total energy insufficient? The energy is ##1.1 \times M_{4s}##?
     
  17. May 16, 2015 #16

    mfb

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    In the lab frame, yes, but producing an Y(4s) at gamma=1.1 would violate momentum conservation.
     
  18. May 18, 2015 #17
    So in the CM frame the total energy is ##M_{4s}## while in the lab frame one has ##1.1 \times E_{-}## energy. I would use invariance of the length of 4-vector to find the energies?
     
  19. May 18, 2015 #18

    mfb

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    Correct.
     
  20. May 18, 2015 #19
    [tex]M^2 = (1+\delta)E_{-}^2 - (p_1-p_2)^2 [/tex]
    [tex]M^2 = (1+\delta)E_{-}^2 - p_1^2 - p_2^2 + 2p_1p_2[/tex]
    [tex]M^2 = \delta E^2(1-\delta) + 2m^2 + \sqrt{\left[ E_{-}^2-m^2\right]\left[ (\delta E_{-})^2-m^2 \right]}[/tex]

    Still can't solve for ##E_{-}##..
     
  21. May 19, 2015 #20

    mfb

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    E is the only unknown quantity in that equation.
    You can neglect the electron mass - 0.0005 GeV are irrelevant. That simplifies the equations a lot.
     
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