# What would happen if you reduced the coupling of $SU(2)$ in the standard model to zero?

Ultimately, my goal is to find a free parameter that you could change in order to significantly reduce the strength of, or eliminate, the weak interaction. Would such a modification leave other parts of the Standard Model unchanged?

• This question (v2) seems quite broad. Aug 4 at 6:15

It is straightforward to see, even though your ultimate vision should be in trouble. I assume you mean decrease the coupling g of just SU(2), and leave the EM coupling e and the Higgs v.e.v. v alone, which cannot be done.

You then just look at the formulas: $$\cos \theta_\text{W} = \frac{g}{\,\sqrt{g^2+g'^2\,}\,}, \qquad \sin\theta_\text{W} = \frac{g'}{\,\sqrt{g^2+g'^2\,}\,} \\ e=g\sin\theta_W= g'\cos \theta_W\\ m_\text{Z} = \frac{m_\text{W}}{\,\cos\theta_\text{W}\,}, \qquad m_W= {ev\over 2\sin\theta_W}, \qquad G_F= 1/(v^2\sqrt{2}).$$

As $$g\to 0$$, the Weinberg angle increases to π/2, its cosine vanishes, and its sine goes to 1.

• But note, the EM charge e cannot stay invariant, since $$e\to g$$.

The Fermi constant stays put; the mass of the W goes to zero (it stops coupling!), and the mass of the Z goes to $$g'v/2$$.

NB aside: If you are seeking decoupling, the opposite limit, $$g\to \infty$$, paradoxically is better behaved, and often taught in class: in that case, e can stay unchanged, $$e=g'$$, since $$\theta_W=0$$, the cos is 1, and $$m_Z=m_W=\infty$$, while the Fermi constant is what it always was: the old Fermi theory! So you may think of EW unification as a descent from infinite to a finite g...

• Wait - I have a question. In your postscript, are you saying that you can eliminate the weak interaction by increasing $g$ to $\infty$? What does it mean to have bosons with infinite mass? Perhaps it would be also interesting to explore a more intermediate case - what if $g$ was significantly larger? Aug 4 at 17:26