# What's the most precise test of electroweak unification in the standard model?

As I understood, there is a coincidence of the weak coupling constant $g$ calculated in two different ways:

1) The muon lifetime $\tau_{\mu}$ is related to $g$ by the formula $(m_{\mu} c^2)^5 \tau_{\mu}= (8 \pi)^3 12 \hbar (\frac{m_W c^2}{g})^4$, where $m_{\mu}$ and $m_W$ are the masses of the muon and the W-boson. Putting in $m_W= 80.385 \frac{GeV}{c^2}$, $m_{\mu}= 105.65837 \frac{MeV}{c^2}$ and $\tau_{\mu} = 2.1969811 \ ms$ yields $g=0.65224$.

2) If for the Weinberg angle $\theta_W$ holds $\cos \theta_W = \frac{m_W}{m_Z}$, $\sin \theta_W = \frac{\sqrt{4 \pi \alpha}}{g}$, then with $1/\alpha = 137.035999$, and the Z boson mass $m_Z =91.1876 \frac{GeV}{c^2}$ it follows $g=0.64144$.

The agreement is about $1.7$ percent. Is this the most precise (quantitative) test of electroweak unification, or are there other tests with a higher precision? (I do $not$ mean tests of QED such as the Lamb shift or the anomalous magnetic momentum of the electron)

• The PDG review (pdf) is always relevant. Further searching the arxiv for precision electroweak constraints brings up many hits. Here are a few of the more recent ones: 1306.0571, 1307.6173, 1306.3380. – Michael Brown Aug 23 '13 at 9:59
• By the way, the precision of electroweak measurements these days is such that you need to include radiative corrections in everything. So the tree level formula for, e.g. the Weinberg angle, doesn't correspond to what you actually measure. – Michael Brown Aug 23 '13 at 10:37