Origin of the value of the vacuum v ~ 246 GeV

In summary, the value of the Higgs vev in the standard model, which is determined by the parameters mu^2 and lambda in the Higgs potential, is also connected to the Fermi constant G_F. This is determined by calculating the W mass and using the decay rate of the muon to eliminate the vev in terms of G_F. This is the most precise method for measuring G_F.
  • #1
Sleuth
47
4
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
 
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  • #2
The Higgs vev is mu/sqrt(lambda), but it's also 2M_W/g. You calculate the W mass and get vg/2.
 
  • #3
Sleuth said:
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
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
 
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Likes ChrisVer
  • #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.
 
  • #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?
 
  • #6
Sleuth said:
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,

If you do that, you recover it with strength zero.
 
  • #7
Yes well, I meant keeping the first non-zero order in 1/m_W of course...
Thanks again
 
  • #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
 
  • #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 :)
 

Related to Origin of the value of the vacuum v ~ 246 GeV

1. What is the significance of the vacuum value v ~ 246 GeV?

The vacuum value v ~ 246 GeV is a crucial parameter in the Standard Model of particle physics. It is the energy scale at which the Higgs field, responsible for giving particles their mass, is believed to have a non-zero value.

2. How was the value of v determined to be approximately 246 GeV?

The value of v was first predicted by theoretical calculations in the 1970s, and was later confirmed by experiments at the Large Electron-Positron Collider (LEP) in the 1990s. More precise measurements have been made at the Large Hadron Collider (LHC) in more recent years, confirming the value to be around 246 GeV.

3. What is the relationship between the vacuum value v and the Higgs boson?

The vacuum value v is directly related to the mass of the Higgs boson. In fact, the mass of the Higgs boson is approximately v times the strength of the Higgs field. This means that a larger v value would result in a heavier Higgs boson.

4. Is the value of v constant or can it change?

The value of v is believed to be a constant in our universe. However, some theories suggest that in the early universe, the value of v may have been different, resulting in a different mass for the Higgs boson and potentially altering the development of our universe.

5. What implications does the value of v have for particle physics and the study of the universe?

The value of v has significant implications for our understanding of the fundamental forces and particles in our universe. It helps to explain how particles acquire mass and how they interact with each other. Further research on the value of v and the Higgs boson could potentially lead to a more comprehensive theory of particle physics and the formation of our universe.

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