Beta Function for QED: 20/9 or 8/3?

[RedX]

The beta function for QED is given by:

\beta=\frac{e^3}{16 \pi^2}*\frac{4}{3}*(Q_i)^2

where (Q_i)^2 represents the sum of the squares of the charges of all Dirac fields.

For one generation, for the charge squared you have (2/3)^2 for the up quark, (-1/3)^2 for the down quark, but this is all multiplied by 3 for the 3 colors of quarks, and then you have (-1)^2 for the electron and (0)^2 for its neutrino.

So all in all, 3[(2/3)^2+(-1/3)^2]+(-1)^2=8/3

However this gives a beta function that is not equal to the book value of:

\beta=\frac{e^3}{16 \pi^2}*\frac{20}{9}

So is the book wrong?

 

[Avodyne]

20/9 is the coefficient for the contribution of one quark-lepton generation to the beta function for hypercharge, not QED.

 

[RedX]

Ah that's it. But now I have to re-calculate the beta function for the hypercharge. Darn it.

Do you remember Srednicki ever providing justification that the coupling constants of SU(3), SU(2), and U(1) are unified to the coupling constant of SU(5) (well within a factor of 3/5 for the U(1) coupling constant) at the mass of the X-particle? I mean you can show that the 3 couplings of the product group meet at a point, but after that point what's the justification for using SU(5) instead of just continuing to follow the three beta functions past that point, i.e., continue with SU(3)xSU(2)xU(1)? I don't recall ever seeing Srednicki talk about the phase transition.

 

[Avodyne]

He says after eq.(97.14) that the couplings are equal in the MS-bar renormalization scheme, but that we should not use this scheme below MX. Then, at the bottom of p.630, he says that the usual beta functions apply if we integrate out the heavy fields. At the top of p.631, he says that we have to restore the heavy fields for mu>MX, and then the couplings are equal again. So it just depends on which renormalization scheme is most appropriate, the one with heavy fields included or the one with heavy fields integrated out.