The Higgs Mass and Spontaneous Symmetry Breaking in Cosmology

In summary: The higgs mass does not depend on temperature. The universe was in an unbroken phase, but that phase ended and the universe is now in a broken phase.
  • #1
kalish
28
0
Hello, I am searching for information about the higgs mass before and after the symetry breaking. Does it depend on the temperature? I know other particles are not expected to have a mass before symetry breaking, but I suppose the higgs itself does. I don't find much informations about and I don't find the involvment in cosmology anymore.

Maybe I don't search well. Could anyone help me?
 
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  • #2
The Higgs potential is usually taken to be V(φ) = μ2φ2 + λφ4 where λ > 0 and μ2 < 0. If you take this to be literally true, then in the initial unbroken phase the Higgs would be tachyonic. No one knows whether the form assumed for V(φ) has any connection with reality other than the fact that there is a minimum.
 
  • #3
Ok so there has never been an unbroken phase in the universe? for example with [itex]\mu^2>0[/itex]

I still quote the paper of sean carroll talking about the cosmological constant:

In the Weinberg-Salam electroweak model, the phases of broken
and unbroken symmetry are distinguished by a potential energy difference of approximately
MEW ∼ 200 GeV (where 1 GeV = 1.6 × 10−3 erg); the universe is in the broken-symmetry
phase during our current low-temperature epoch, and is believed to have been in the symmetric phase at sufficiently high temperatures in the early universe.

I just want to know why do we believe this , and thus the dependence of [itex]\mu[/itex] on T.
I just can't find a clear paper about this.
 
  • #4
μ2 does not depend on T, or anything else. And as I said, μ2 < 0. I said there was an initial unbroken phase. With, guess what: μ2 < 0.
 
  • #5
Hello, do you have some source about that? I am sorry but I very doubt about your claim. The unbroken symetry imply [itex]\mu^2>0[/itex] since the derivative shows that the zero are for [itex]\phi = 0[/itex] or [itex]\phi^2 = -\frac{\mu^2}{4 \lambda}[/itex], it follows when [itex]\mu^2 < 0 [/itex], then [itex]\phi_0 = \pm \sqrt{-\frac{\mu^2}{4 \lambda}}[/itex] has real solutions they are minimum of the potential, and then it is a broken phase.
 
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  • #6
I'll try one more time, kalish. The Higgs potential is not supposed to change. It has a minimum at φ = √(- μ2/4λ), V(φ) = - μ2/4λ, which is about 200 GeV. This point is a stable vacuum state. Under present conditions the Higgs field sits down at the bottom of the valley and the symmetry is broken.

V(φ) also has a maximum at φ = 0, V(φ) = 0. This point is unstable, a false vacuum. For extremely high temperatures, E >> 0 and where is the Higgs field? Everywhere. It is not confined to the valley. In fact for very high temperatures it does not even notice the valley, the details of the potential no longer matter. Certainly the sign of μ2 does not matter. This is why you won't find an answer to your question about the Higgs mass. Rest mass is a property of particles at rest or nearly so. In the early universe the kinetic energy greatly exceeded the current energy scale of 200 GeV, and at that high energy scale by comparison the rest mass is insignificant.
 
  • #7
ok, thanks that's what i wanted to know, indeed at our epoch the higgs is just supposed to get it's vacuum expectation value, and an hypothetical symmetric vev when [itex]\mu^2 > 0[/itex] has never happend? I am sorry for the trouble, I am not a real retard, but I read everywhere the presentation talking about the two values of [itex]\mu^2[/itex] to present what was a spontaneous symmetry breaking, (I can quote!) so I effectively expected this to be a physical process (you know the parabol and the hat). So the point is just now it is almost at the vev, but at earlier time it was just not. It is different than the vev was different earlier.

Thanks a lot you avoid me troubles.
 

Related to The Higgs Mass and Spontaneous Symmetry Breaking in Cosmology

What is the Higgs mass before breaking?

The Higgs mass before breaking refers to the mass of the Higgs boson particle before the Higgs field is activated and breaks the electroweak symmetry.

Why is the Higgs mass before breaking important?

The Higgs mass before breaking is important because it helps explain why particles have mass. Before the Higgs field breaks, particles are massless and can move at the speed of light. After the breaking, particles gain mass and slow down, allowing for the formation of atoms and ultimately, the universe as we know it.

How is the Higgs mass before breaking calculated?

The Higgs mass before breaking is calculated using the Standard Model of particle physics, which predicts the mass of the Higgs boson based on the strength of the Higgs field and other fundamental constants. Scientists also use particle accelerators, such as the Large Hadron Collider, to directly measure the mass of the Higgs boson.

What is the relationship between the Higgs mass before breaking and the Higgs field?

The Higgs mass before breaking and the Higgs field are closely related. The Higgs field is a theoretical field that permeates the universe, and its strength determines the mass of particles. The Higgs boson is the particle associated with the Higgs field and its mass is a reflection of the strength of the field.

Can the Higgs mass before breaking change?

The Higgs mass before breaking is considered a fundamental constant in the Standard Model of particle physics, meaning it is thought to be unchanging. However, some theories suggest the existence of multiple Higgs fields and bosons, which could potentially lead to variations in the Higgs mass before breaking. Ongoing research and experimentation are needed to fully understand the nature of the Higgs boson and its mass.

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