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Origin of the value of the vacuum v ~ 246 GeV

  1. Dec 19, 2015 #1
    Hi guys,
    I have a very very simple and naive question, which I hope I shouldn't be ashamed to ask, but I would like to clarify an issue in my mind.

    We hear everywhere that the value of the ew vacuum is v ~ 246GeV, fixed by the Fermi Constant G_F, v = (\sqrt(2)*G_F)^(-1/2).

    Now this is also the vacuum of the Higgs potential in the standard model, which is determined by the two parameters, say mu^2 and \lambda, in the Higgs potential, schematically

    V(\phi) = mu^2 phi^2 + \lambda \phi^4.

    The mechanism does not have anything to do with G_F at this point, it exists completely independently of the rest of the Lagrangian which can contain any other interactions.

    Therefore the question is, at which step of the implementation of the mechanism in the EW SU(2)xU(1) and how exactly do we get to say that this v is actually connected to G_F, and in particular it is exactly equal to (\sqrt(2)*G_F)^(-1/2)? Does one necessarily need to compute, for example, the process

    \mu -> \nu + \bar{\nu} + e^+

    in the SM in terms of the parameter v, then take the limit M_W->\infty, get the corresponding "effective" 4-point interaction and then compare it to what Fermi had called G_F, or can one somehow make this statement"a priori" from more general considerations?

    Also, once one agrees on this, what is the most precise method to measure G_F?

    Thanks guys and sorry if this sounds a bit silly...
    Sleuth
     
  2. jcsd
  3. Dec 19, 2015 #2

    Vanadium 50

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    The Higgs vev is mu/sqrt(lambda), but it's also 2M_W/g. You calculate the W mass and get vg/2.
     
  4. Dec 19, 2015 #3

    samalkhaiat

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    See the last part (equations 36-40) of the post
    https://www.physicsforums.com/threads/why-su-2-times-u-1-for-the-sm.846099/#post-5320206
     
  5. Dec 20, 2015 #4
    One doesn't really send mw-> infinity, but rather ignore the electron mass and also the momentum transfer in muon decay. Then you get a decay rate proportional to 1/vev^2 from muon decay (as people point out related to the W boson coupling to a lepton su2 doublet). Then this decay rate is used to eliminate the vev in terms of gf as measured from muon decay.

    This is the general procedure. To be more accurate, people actually generally consider the one loop corrections to muon decay, and eliminate the vev this way.
     
  6. Dec 20, 2015 #5
    Thanks for the answers! Sending m_W to infinity would be the sensible thing to do for me, if one wants to recover the Fermi 4-point like interaction, but I guess this is only matter of language.

    Does anyone know exactly what is the best measurement of G_F (and therefore of v)? I guess one could measures the muon decay rate and the muon, or is there anything better?
     
  7. Dec 20, 2015 #6

    Vanadium 50

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    If you do that, you recover it with strength zero.
     
  8. Dec 21, 2015 #7
    Yes well, I meant keeping the first non-zero order in 1/m_W of course....
    Thanks again
     
  9. Dec 22, 2015 #8
    ;) I knew what you meant, but just thought it was sloppy terminology.

    For sure, just drop the momentum dependence of the interaction.

    Muon decay is the most precisely measured experimental process for extracting and vev. People have also been computing corrections to this process (weak, qed, effective field theory corrections etc.) so it's also theoretically well motivated to use it.

    Note that this is just a choice of input parameters. One could just quite as well use m_w , electric charge, m_z. It's just that m_w is a couple of orders of magnitude less precisely known
     
  10. Dec 22, 2015 #9
    Thanks RGevo. Yes I was being a bit sloppy, but I thought that was clear enough.

    I wanted to make sure I knew what was going on regarding the value of v. The discussion helped clarifying my mind :)
     
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