The W is a very unbiased particle - it decays into any two particles it can with equal probability. This decay is always to a pair of particles, either quarks or leptons, and it must conserve charge. As the W has charge either + or - 1, so must the total of the decay products. Also, the decay products must weigh less than the W they started from, or energy would not be conserved. This excludes the top quark. This means that the following decays are possible:
| W- | W+ |
|---|---|
| down and anti-up quarks | anti-down and up quarks |
| strange and anti-charm quarks | anti-strange and charm quarks |
| electron and anti-neutrino | positron and neutrino |
| muon and anti-neutrino | anti-muon and neutrino |
| tau lepton and anti-neutrino | anti-tau lepton and neutrino |
The reaction e+e- to W+ does not conserve charge, and is not possible. Instead we use e+e- to W+W- . This means that we have a pair of W's in the same event to understand, and the events are more complicated than Z decays.
| W+ decay | W- decay | Probablity IF only 1 colour |
|---|---|---|
| Two jets from quarks | Two jets from quarks | 2/5 * 2/5 = 16% |
| Two jets from quarks | lepton and neutrino | 2/5 * 3/5 = 24% |
| lepton and neutrino | Two jets from quarks | 3/5 * 2/5 = 24% |
| lepton and neutrino | lepton and neutrino | 3/5 * 3/5 = 36% |
In general we can see that if there are ncolours colours there are 2ncolours quark modes for each W and 3 leptonic ones. As the W's decay equally to all modes, if we get a large sample of W pairs we will find the relative proprtions of events given in the following table:
| leptonic | Mixed | Four jets |
|---|---|---|
| l | m | f |
| 9 | 12n | 4n2 |
If we take the numbers of events in any two of the categories f,l and m we can solve the equation to obtain ncolours. So we have 3 ratios we can use:
It turns out that the last method is the most accurate, so we will use that. (But the middle one is almost as good - you might like to compare)