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A Weinberg angle in terms of Higgs mass

  1. Feb 21, 2017 #1
    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. jcsd
  3. Feb 21, 2017 #2

    Vanadium 50

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    Of course not. Did you even try and plug the numbers in?
     
  4. Feb 21, 2017 #3

    ChrisVer

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    why?
    that's the sin^2:
    [itex]\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 [/itex]
    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 [itex]v[/itex]) are found:
    [itex] M_W = v \frac{g}{2}[/itex]
    [itex] M_Z = v \frac{\sqrt{g^2+g'^2}}{2}[/itex]
    [itex] \cos \theta \equiv \frac{M_Z}{M_W} = \frac{g}{\sqrt{g^2+g'^2}}[/itex]
    with [itex]g,g'[/itex] the coupling constants of [itex]SU(2),U(1)[/itex] respectively. This is only a function of the coupling constants of [itex]SU(2)[/itex] and [itex]U(1)[/itex] (can be interpreted as the angle between the constants if you represent them as orthogonal vectors). If [itex]\cos \theta[/itex] is not a function of any quantity that's proportional to the Higgs' mass ([itex]m_H^2 =2 v^2 \lambda [/itex]), then neither is [itex]\theta[/itex].
     
    Last edited: Feb 21, 2017
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