# A Explanation of terminology: electroweak

1. Mar 15, 2016

### dextercioby

I'm not a specialist in this subject, so bear with me. I've always wondered why one claims that the electromagnetic and weak interactions are unified, but the strong one with the (unified) other two is not.
Mathematically, I'm aware that the full gauge group of the SM is $U(1) \times SU(2) \times SU(3)$, so one perceives all three interactions separately and on equal footing. In what exact sense are the weak and the em unified, but the strong not?

Thank you!

2. Mar 15, 2016

Staff Emeritus
Since this is an A, what do you think the U(1) and SU(2) are?

3. Mar 15, 2016

### nrqed

U(1) has one generator and SU(2) has three generators. The key questions are: The E&M gauge field is associated to which of these generators? The $Z_0$ is associated to which ones? What about the $W^\pm$?

4. Mar 15, 2016

### dextercioby

U(1) is the gauge group of electromagnetism, while SU(2)w is the gauge group of the weak interactions. The fields are all 4 vectors (co-vectors actually): the e-m potential A and Wa, a=1,2,3. SU(2) (just as SU(3) in QCD) enters through the adjoint representation of dimension 3.

One more time, why one claims that A and W are "unified", if the there's no single coupling constant and no global compact and connected gauge group which has the direct product U(1) x SU (2) as a subgroup?

5. Mar 15, 2016

### nrqed

Actually, U(1) is not the gauge group of electromagnetism and SU(2) is not the gauge group of the weak interaction. That was the point I wanted to make. That U(1) is the weak hypercharge, not the electromagnetic U(1). What happens is that electromagnetism corresponds to a linear combination of the weak hypercharge generator and of the $T_3$ diagonal generator of the weak isospin SU(2), so that the electric charge is given by $Q = T_3 + Y/2$ where (by abuse of notation) here $T_3$ is the eigenvalue of the diagonal generator of the weak isospin SU(2), and Y is the hypercharge. The orthogonal linear combination corresponds to the $Z_0$ boson while the $T_\pm$ correspond to the $W^\pm$. It is in that sense that the weak interaction and the electromagnetic force are deeply linked. Now, you are right that they do not not have the same coupling constant since the coupling constants of the hypercharge U(1) and the weak isospin are not equal and this shows up through the Weinberg angle, so I agree with you that saying they are "unified" is a stretch. But they are definitely deeply linked together, in the sense I just described.

6. Mar 15, 2016

### dextercioby

Thanks for pointing that essential aspect to me. Now I understand.

7. Mar 15, 2016

### samalkhaiat

8. Mar 16, 2016