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?

  • 2
    $\begingroup$ This question (v2) seems quite broad. $\endgroup$
    – Qmechanic
    Aug 4 at 6:15

1 Answer 1


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

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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...

  • $\begingroup$ Thank for your answer! $\endgroup$
    – edzatle
    Aug 4 at 9:01
  • $\begingroup$ 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? $\endgroup$
    – edzatle
    Aug 4 at 17:26
  • $\begingroup$ Not eliminate! Hide the gauge bosons thereof! The Fermi constant stays the same, so the four-fermion interaction is still what it was for 40 years (1933-1973), when they didn't quite know how heavy the vector bosons were! Plug g s twice, 10 times, 100 times the real value! $\endgroup$ Aug 4 at 17:37
  • $\begingroup$ Wouldn't the weak force still exist where you have virtual particles mediating the interactions? Although a temporary disturbance in the quantum field with infinite mass is crazy to think about... $\endgroup$
    – edzatle
    Aug 4 at 18:26
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    $\begingroup$ Ahhh, okay i see $\endgroup$
    – edzatle
    Aug 4 at 19:36

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