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How many W mesons required?

  1. Nov 16, 2007 #1
    1. The problem statement, all variables and given/known data
    The decay
    [tex]D^{+} \rightarrow D^{0} \pi^{+}[/tex]
    is not possible via a weak interaction which involves one virtual W+ meson. How many virtual W mesons are required?

    2. Relevant equations
    D0 has quark content c-ubar
    Pi+ has quark content u-dbar
    I'm not sure what D+ has...

    3. The attempt at a solution
    Assuming D+ has c-dbar...
    how do you determine whether a virtual meson is involved or not and how many? Is it energy based?
  2. jcsd
  3. Nov 16, 2007 #2


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    Last edited: Nov 16, 2007
  4. Nov 16, 2007 #3
    thanks very much!
    so D+ has c-dbar...
    D+ = 1869MeV
    D- = 1865MeV
    pi+ = 139.6MeV

    So that means both a W+ and a W- boson are required to keep the charge at +1 on both sides? And have energy greater than 139.6-5?
  5. Nov 16, 2007 #4


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    Even though you had more W bosons involved, i say that it violates energy.
  6. Nov 16, 2007 #5
    okay, thanks... so a trick question? An even number of bosons would be required to conserve charge, but energy will never be conserved so it can't happen...
  7. Nov 17, 2007 #6


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    Hmm ONE W+ boson also conserve charge.

    In each vertex, charge, lepton number etc are conserved.

    imagine the following:
    In the D+ , the W+ turns the d-bar quark to an u-bar quark, and latter decays into something.

    The thing is that you have less mass on the RHS of the eq, W+ can decay into one pion + meson. so there is no problem besides energy.

    I THINK that you have made a misstake, that it should be the [tex] D^x _S [/tex] Meson instead, then you have more mass (maybe correct mass for it to be allowed energetically). I might be wrong, but that is one desperate way =)
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