How many W mesons required for D+ decay?

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Homework Help Overview

The discussion revolves around the decay process of the D+ meson, specifically the reaction D^{+} \rightarrow D^{0} \pi^{+}. Participants are exploring the implications of weak interactions and the role of virtual W mesons in this decay process.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to identify the quark content of the D+ meson and its implications for the decay process. Questions are raised about the number of virtual W mesons involved and how energy conservation plays a role in determining this. Some participants suggest using Feynman diagrams to visualize the interactions.

Discussion Status

The discussion is ongoing, with participants sharing insights about the quark content and energy considerations. Some guidance has been offered regarding the conservation of charge and the potential need for multiple W bosons, but there is no explicit consensus on the number required or the feasibility of the decay.

Contextual Notes

Participants are navigating the complexities of particle decay and conservation laws, with some uncertainty about the energy balance and the specific mesons involved. There are references to external resources for further clarification on meson properties and W boson behavior.

genloz
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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?
 
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Last edited:
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?
 
Even though you had more W bosons involved, i say that it violates energy.
 
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...
 
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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