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Gauge bosons and the weak mixing angle

  1. Aug 20, 2015 #1
    <<Mentor note: Moved from other thread.>>

    I have 4 questions:

    1. Why Weinberg angle affects neutral boson mixing, while W+ and W- are unaffected?
    2. Is there any relation between Weinberg angle and CP violation angle? Are they absolutely independent?
    3. How our world would be different if value of the angle was different, especially in extreme cases 0 and 90?
    4. Z appears only slightly different from photon, why Higgs boson interacts with Z but not with photon? They both "contain" W3 and B!!! How Higgs interacts with W3 and B before symmetry breaking? (math of electroweak theory is far beyond me, this is why I am asking, may be there is a simple explanation)

    Thank you
     
    Last edited by a moderator: Aug 20, 2015
  2. jcsd
  3. Aug 20, 2015 #2

    ChrisVer

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    Because there's a mixing with the SU(2) and U(1) that are broken, and the mixing angle between those gauge groups is the Weinberg angle. Don't forget that the weinberg angle can be written in term of the coupling constant characteristic of each of those two groups divided by a "distance" of them.
    [itex] \cos \theta_w = \frac{g_2}{\sqrt{g_2^2 + g_Y^2}} ~~, ~~\sin \theta_w = \frac{g_Y}{\sqrt{g_2^2+g_Y^2}}[/itex]
    You can build triangles with g_Y and g_2 and illustrate it.

    I am not sure about it, but I think there isn't any relation. They are independent free parameters of the SM.

    One thing I see from just looking at the expressions of Z,gamma is that there will be no interference between them.
    However let's go back again at the cos,sin expressions. If theta was 0 or the cos=1, then obviously you get that you didn't have any U(1) contribution in the first place [itex]g_Y \rightarrow 0[/itex]. If that was the case then the masses of the W and Z would be equal because:
    [itex]m_W^2 = \frac{g_2^2 }{4} v^2_F ~~,~~ m_Z^2 = \frac{g_2^2 + g_Y^2}{4}v_F^2[/itex]
    or
    [itex] \frac{m_W^2}{m_Z^2} =\frac{g_2^2 }{g_Y^2+g_2^2} = \cos^2 \theta_w [/itex]

    In a similar manner you get the same for SU(2) with π/2 ([itex]g_2 \rightarrow 0[/itex]). In that case, strangely the W mass appears to be very small/almost zero while the Z mass will not.

    I think this is quite natural in any case, since by sending g_Y or g_2 to 0, you are not breaking that part of the symmetry.
    So for example in cos=1 case the U(1)_Y remains unbroken (and so there is hardly any "mixing" coming from it in SU(2) ).
    For the cos=0 case, the SU(2) remains unbroken and U(1) breaks. The massive boson is then only 1, while you have 3 remaining massless bosons: Ws and photon.

    These are a few "conclusions" I reached by looking at some formulas.

    The maths are quiet clear and you can find the answer by working 'em out.
    A fast answer good for almost anyone who has seen the Higgs potential, is because there is still a flat direction at the Higgs potential after the Higgs field acqured a vacuum expectation value breaking spontaneously the SU(2)xU(1) into a single U(1) (so you still have to have a massless boson). Imagine the bottom of the mexican hat, where you can rotate around freely (that's the remained U(1) transformation of the Higgs field)
     
    Last edited: Aug 20, 2015
  4. Aug 20, 2015 #3
    Thank you for detailed answer!

    ... to be more specific, how sensitive life (life-friendly laws of our universe) is to that angle?
    I mean, does changing it slightly break something completely in, say, creating/stability of hadrons, atoms, nuclei etc, or you can change with it without dramatic consequences say in ranges 5-70 degrees?
     
  5. Aug 20, 2015 #4

    ChrisVer

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    if you change it, you are obviously changing the masses of the W,Z bosons [their ratio]. So yes, in fact you are changing the physics (I cannot answer reliably on that, since it wouldn't make sense).
    However it's not a constant by itself, since the coupling constants change with energy (but I think that's not what you are asking).
    This is off-topic in general, if you like create another thread.
     
  6. Aug 20, 2015 #5
    Why it wouldn't make sense?
    It makes a perfect sense in, eternal-inflation-with-baby-universes-with-different-parameters-of-standard-model scenario.
    Sorry for off topic, but in general, I would like to understand what is a "shape" of a life-friendly cloud in 20+ dimensional space, and how close are we to the center However, I've never seen any articles regarding this, where all parameters are analyzed. Of course, we all know about "lucky coincidences" like Carbon cycle, absence of nuclei with 5 and 8 hadrons, neutron "almost stability" etc, but I've never seen such things mapped directly into parameters of the standard model.
     
  7. Aug 20, 2015 #6

    Orodruin

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    In the Higgs mechanism, there is a residual U(1) symmetry left after symmetry breaking. The photon corresponds to this symmetry and is therefore massless and does not interact with the Higgs field. Said in a different way, the photon is defined as the electrically neutral gauge boson which does not have Higgs field interactions.

    As has already been indicated, the Weinberg angle is ultimately dependent on the coupling strengths of the gauge groups. If ##\sin(\theta) = 0##, then the hypercharge gauge group coupling is zero and if ##\cos(\theta) = 0##, then the SU(2) gauge coupling is zero. In the former case, the hypercharge does not interact with anything. The Higgs interaction with the gauge bosons becomes ##g^2 W^2 \Phi^2##, resulting in an equal mass for all SU(2) gauge bosons. The hypercharge boson remains massless (and completely decoupled from the rest of the world, you might just leave it out of your theory).

    When ##\cos(\theta) = 0##, the SU(2) does not couple at all and the ##Z## is equal to the hypercharge gauge boson, which will obtain a mass from the fact that the Higgs also has a hypercharge. The Higgs being an SU(2) doublet becomes meaningless as the SU(2) does not couple at all. You are left with three rather uninteresting non-interacting ##W## fields (which you might just leave out of your theory).
     
  8. Aug 20, 2015 #7

    ChrisVer

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    Because changing the physics of something, in this case the masses of the heavy bosons (W,Z) by playing around with the Weinberg angle, is going to change many things about the weak interactions (unobserved changes => unreal)...
    As for if this can have an effect on how the universe evolved (for example some particle decay lifetimes will be altered). How will they alter I am in no position to answer (depends).


    PS to mentors : my post #2 here was an answer to a post tht exists in the other thread :DD
     
  9. Aug 20, 2015 #8

    Orodruin

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    I am not sure I understand what you are talking about ... :rolleyes::rolleyes::rolleyes::-p
     
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