How to determine quantum numbers for beta functions?

[lonewolf219]

I'm trying to understand the notation (3, 1, 2/3) for the up quark and (3, 2, 1/6) for the left-handed up and down quarks... Is the first number related to SU(3), the second SU(2) and the third I believe is the hyper charge... Not sure what the significance is of the first two numbers...

I think the 1 in (3, 1, 2/3) means the up quark doesn't interact with the weak force. But what would the 3 mean? A triplet in SU(3)? If so, how would I find it?

Possible equations:

Q = I3 + Y
Q = I3 - Y
Q = T3 + Y

 

[Bill_K]

The 3 means the quarks are members of a color triplet. Otherwise they are characterized by weak isospin T and weak hypercharge Y. The relationship connecting T and Y to Q is Q = T₃ + Y/2.

Left-handed fermions are weak isodoublets. Right-handed ones are weak isosinglets. This means they don't interact with the W boson, but they still do interact with the Z. Quark values:

u_L: T = 1/2, T₃ = +1/2, Y = 1/3, Q = 2/3
d_L: T = 1/2, T₃ = -1/2, Y = 1/3, Q = -1/3
u_R: T = 0, T₃ = 0, Y = 4/3, Q = 2/3
d_R: T = 0. T₃ = 0, Y = -2/3, Q = -1/3

 

[lonewolf219]

OK, that makes sense. Thank you, Bill K!

...Do right-handed quarks couple to the W or Z bosons? My guess is no, which might be why there is not a right-handed quark doublet? Is this because their weak isospin is 0?

 

[lonewolf219]

Can anyone tell me what this process is called, so maybe I can read a bit more about it? Any suggestions would be great

 

[Avodyne]

The relationship connecting T and Y to Q is Q = T₃ + Y/2.

Some authors (e.g., Srednicki) normalize hypercharge so that Q = T₃ + Y. In the OP's notation of (3,1,2/3) for the right-handed up quark, this is the normalization that is used.

For the basics of how the various fields interact, see "After electroweak symmetry breaking" in
http://en.wikipedia.org/wiki/Electroweak_interaction

For more details, see any good book on particle physics or QFT of the Standard Model.

 

[lonewolf219]

Thank you, Avodyne... Helpful! Thanks for pointing out it's normalized... I don't think I could have figured that out on my own :)