Help! Quick Question on Calculating Quark Jet Branching Ratios

[venomxx]

Quick question - Help!

Im working on branching ratios of quark jets, the book is pulling fractions out of nowhere and I am wondering how its getting them!

for example with a beam of e+ and e- colliding you get quark jets, the book says the ratio of the crossections of the collsions going to dd* compared to muons (where* designates the anti-particle) is (1/3)^2 and the ratio given for uu* created is (2/3)^2.

Therefore total = 5/9

The notation the book uses is σ(ee --> dd*)/σ(ee -->mu mu) = (1/3)^2


As the centre of mass energy increases other heavier quarks can become possible...

Can anyone shed light on where these fractions come from? I am sure there easily calculated.

 

[ansgar]

quarks carry different charges than charged leptons -> This changes the magnitude of the coupling-strength

also please use better names for your topics "help" is not a good title...

 

[venomxx]

Cheers, i won't use that title again.

Am i right in saying that Up, Charm and Top quarks have 2/3 charge and Down Strange and bottom have -1/3 charge so this is where these fractions come from - the relationship you mentioned with coupling constants and charge?

If that is the case, you say leptons are different so would they have to be treated differently?

 

[ansgar]

Cheers, i won't use that title again.

Am i right in saying that Up, Charm and Top quarks have 2/3 charge and Down Strange and bottom have -1/3 charge so this is where these fractions come from - the relationship you mentioned with coupling constants and charge?

If that is the case, you say leptons are different so would they have to be treated differently?

The electromagnetic force couples to the product of charges of the particles in the interaction vertex, so leptons have integer charges and quarks 3rds of integers therefore you must treat them "different"

You have neglect one thing, that quarks have one additional quantum number, the color, which affects the cross-section (not the interaction itself) by a factor of 3.

The probability to create a red-antired d,dbar pair is the same as a green-antigreend,dbar pair, and so on.