# A Weinberg angle in terms of Higgs mass

1. Feb 21, 2017

### spaghetti3451

The Weinberg angle $\Theta_{W}$ is commonly expressed as

$$\cos\Theta_{W} = M_{W}/M_{Z}.$$

Can the Weinberg angle $\Theta_{W}$ be expressed in terms of the Higgs mass and the mass of the W boson as

$$\sin^{2}\Theta_{W}= m_{H}/M_{W}?$$

2. Feb 21, 2017

Staff Emeritus
Of course not. Did you even try and plug the numbers in?

3. Feb 21, 2017

### ChrisVer

why?
that's the sin^2:
$\sin^2 \theta =1 - \cos^2 \theta = 1 - \frac{M_W^2}{M_Z^2} = \frac{M_Z^2 - M_W^2 }{M_Z^2} \approx 0.22$
Your expression doesn't even make sense (for a sin), since the right hand side of your equation is >1 (the higgs is heavier than the W).

And you can't write it in terms of the Higgs mass, because the masses of the W and Z from the Higgs mechanism (with Higgs' vev $v$) are found:
$M_W = v \frac{g}{2}$
$M_Z = v \frac{\sqrt{g^2+g'^2}}{2}$
$\cos \theta \equiv \frac{M_Z}{M_W} = \frac{g}{\sqrt{g^2+g'^2}}$
with $g,g'$ the coupling constants of $SU(2),U(1)$ respectively. This is only a function of the coupling constants of $SU(2)$ and $U(1)$ (can be interpreted as the angle between the constants if you represent them as orthogonal vectors). If $\cos \theta$ is not a function of any quantity that's proportional to the Higgs' mass ($m_H^2 =2 v^2 \lambda$), then neither is $\theta$.

Last edited: Feb 21, 2017