W+ decay modes neglect mass branching ratiosQ

[binbagsss]

Homework Statement

The question is that the decay modes for the $W+$ boson are , $e^+ v_{e}, \mu^+ v_{\mu}, \tau^+ v_{\tau}, ud', cs'$, where a ' denotes a anitquark, neglecting the masses of the decay products estimate the branching ratios of the $W+$?

Homework Equations

N/A

The Attempt at a Solution

I know that the lepton decays differ to quark decays in the way that quarks come in 3 colours.

I however thought that the quark decays would be reduced, as a pose to more likely, than the lepton decays by a factor of $3$.

My reasoning is that , (I believe you treat the 2 quark system as a meson and so the colours must add to neutral, I don't know whether the situation differs if the quarks are unbound or you should consider them unbound in this decay, if someone could answer this, thanks?) , if one of the quarks takes a particular colour, the colour of the other antiquark,say, is immediatey constrained by this as it must be of the same colour, so there is only a $1/3$ chance that it gets the right colour.

Thanks.

 

[mfb]

At this energy, the quarks can be considered as unbound. They will hadronize later, at a lower energy scale.

if one of the quarks takes a particular colour, the colour of the other antiquark,say, is immediatey constrained by this as it must be of the same colour, so there is only a 1/3 chance that it gets the right colour.

You can treat the different quark colors as different particles. Then this constraint is equivalent to "if one particle is a positron, then the other particle has to be an electronneutrino" (as opposed to other neutrinos, or quarks, or whatever) - nothing that would make the decay less likely.
The colors just give three different decay modes: u green d' antigreen, u red d' antired, u blue d' antiblue.

 

[binbagsss]

At this energy, the quarks can be considered as unbound. They will hadronize later, at a lower energy scale.

.

Thanks. I see here as the W boson is colourless so the two quarks colour must sum to zero irrespective of whether it is a meson or not, in the general case of an unbound quark system the colour does not need to sum to zero?

 

[mfb]

I see here as the W boson is colourless so the two quarks colour must sum to zero irrespective of whether it is a meson or not

Right.

in the general case of an unbound quark system the colour does not need to sum to zero?

Right. Gluon+gluon -> quark+antiquark can lead to different colors, for example.