# How many W mesons required?

1. Nov 16, 2007

### genloz

1. The problem statement, all variables and given/known data
The decay
$$D^{+} \rightarrow D^{0} \pi^{+}$$
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. Nov 16, 2007

### malawi_glenn

Last edited: Nov 16, 2007
3. Nov 16, 2007

### genloz

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. Nov 16, 2007

### malawi_glenn

Even though you had more W bosons involved, i say that it violates energy.

5. Nov 16, 2007

### genloz

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. Nov 17, 2007

### malawi_glenn

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 $$D^x _S$$ 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 =)