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How many W mesons required?

  • Thread starter genloz
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  • #1
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Homework Statement


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?

Homework Equations


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

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?
 

Answers and Replies

  • #2
malawi_glenn
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Last edited:
  • #3
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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?
 
  • #4
malawi_glenn
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Even though you had more W bosons involved, i say that it violates energy.
 
  • #5
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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...
 
  • #6
malawi_glenn
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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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