Thanks for any help,
Warren Wolfe
I think that one way to think of it is that the incoming d is more or less
equivalent to an outgoing anti-d.
Another way to think of it might be that u + anti-d -> W is a weak-force
annihilation process, analogous to the usual electromagnetic annihilation
process, in which a photon comes out. In both cases, two fermions come in
and a vector (gauge) boson comes out.
--
Jon Bell <jtbe...@presby.edu> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA
> Can anyone tell me the specific quark interactions that take place in pion
> decay? The end products are all leptons (a muon and a neutrino) so what
> becomes of the up and anti-down quarks? My guess is something like u -> d +
> W(+) and the d annihilates with the anti-d.
You don't need the extra step. The incoming dbar is equavalent to an
outgoing d. u+ dbar -> W+ -> mu+ + nu_mu as described here:
http://hyperphysics.phy-astr.gsu.edu/hbase/particles/piondec.html
> The W(+) can decay to a muon
> plus a neutrino, but what are the by products of the annihilation process?
The muon and the neutrino *are* the byproducts of the decay process.
-E