# Running coupling of the weak interactions

## Main Question or Discussion Point

Can someone point me to a reference that explains how the (effective) running coupling behaves in the weak interactions(at 1-loop order). I couldn't find it...

If I understand correctly, than the coupling is $$g = \frac{e}{sin(\theta_W)}$$ where e is the QED coupling, which increases with energy scale, and $$cos \theta_W = M_W / M_Z$$ is the Weinberg angle. The vector boson masses are in turn determined by their couplings with the Higgs, and the renormalization group equations will determine the running of these (and other) couplings. Than from that we can get the running of this effective coupling. Can someone tell me where to find the results for such a computation? Is it approximately constant until the electroweak unification or not? If not, how much does it change, etc...

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blechman
You can calculate the running of g the same way you calculate the running of g_s. The beta function is always the same:

$$\frac{dg}{d\log\mu}=-\frac{b_0g^3}{16\pi^2}+\mathcal{O}(g^5)\quad{\rm where}\quad b_0=\frac{11N-n_f-n_s}{3}$$

with N=2,3 for g,g_s, and $n_f$ is the number of CHIRAL fermions (so for QCD which has VECTOR fermions, this would be $2n_f$) and $n_s$ is the number of scalars (Higgs bosons).

The g' running (hypercharge) is given by a similar formula, with N=0 (no gauge boson loops) and $n_f\rightarrow 2\sum_f Y_f^2$ where you sum over all fermions, and $Y_f$ is the hypercharge of the fermion (in units of g'). This is the same formula in QED: there are TWO fermions (electron and positron), both with Q^2=1, so that $b_0(QED)=-4/3$. Plugging that into the formula gives

$$\frac{de}{d\log\mu}=+\frac{e^3}{12\pi^2}+\mathcal{O}(e^5)$$

which is correct.

Hope that helps!

Well I am sure that would be the case if the weak interactions was a pure SU(2) gauge theory, but the full SM is more complicated, and I am not sure this still applies.

blechman
sure it does, you just have to be careful to count the fields correctly. For example, when counting quarks in SU(2), remember that you count each quark color as an extra (Weyl) field, so $\Delta n_f=3~({\rm color})\times 3~({\rm generations})=9$ in my formula above (you also have to add leptons). Besides that, it's exactly the same.