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Why Eat Goldstone Boson?

by waterfall
Tags: boson, goldstone
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waterfall
#1
Feb25-12, 06:51 AM
P: 381
I've been reading this very good new book "The Infinity Puzzle" by Frank Close and it mentioned a lot about the Higgs boson eating the Goldstone Boson to have mass. I already checked out wikipedia but can't get an intuitive feel of it. Can anyone explain in a brief mathematical sense why the goldstone boson which has no mass has to be eaten by the higgs boson to make it massive... It would have been more intuitive if the goldstone boson is massive.. and the higgs boson eating it would turn it massive.. but the goldstone is massless. Thanks.
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tom.stoer
#2
Feb25-12, 07:20 AM
Sci Advisor
P: 5,364
Have you ever seen the mexican hat potential?

http://en.wikipedia.org/wiki/Mexican..._hat_potential
http://en.wikipedia.org/wiki/File:Me...tial_polar.svg

Ok, let's first discuss the situation w/o gauge fields

The Goldstone symmetry breaking mechanism says that the rotational symmetry is broken, i.e. that the state is not a V=0 but at V=Vo at some arbitrary angle θo. Now this angle θo breaks the rotational symmetry.

OK, the Goldstone bosons are nothing alese but field fluctutaions at constant V=Vo but fluctuating angle θ(x); b/c the potential is constant in θ-direction there is no force and therefore the fluctuations correspond to massless particles.

Now let's add gauge fields

That means that we do have a local symmetry where at every spacetime point x a local rotational symmetry in θ is present; that means that fluctuations in θ(x) do no longer correspond to particles but to gauge transformations and are therefore unphysical; any change in θ(x) at some spacetime point x can be rotated away by a gauge transformation. That's the reason why Goldstone bosons disappear.

Now back to Vo

http://en.wikipedia.org/wiki/Higgs_mechanism

The radial fluctuations in V still correspond to physical fluctuations, i.e. to Higgs bosons. But instead of having a fluctuation around V=0 we have one around V=Vo. This Vo corresponds to a mass term of the gauge fields (to understand that one has to look into the mathematical details).

The interesting thing is the following: massless gauge fields always correspond to two physical polarizations (two transversal polarizations for el.-mag. waves i.e. two d.o.f. for the photon), but massive vector fields have three degrees of freedom. So what happens is that one degree of freedom disappears from the Higgs field (the θ fluctuation) and reappers as the third (longitudinal) poolarization of the gauge field.

the gauge field eats the massless Goldstone boson
waterfall
#3
Feb25-12, 06:13 PM
P: 381
Quote Quote by tom.stoer View Post
Have you ever seen the mexican hat potential?

http://en.wikipedia.org/wiki/Mexican..._hat_potential
http://en.wikipedia.org/wiki/File:Me...tial_polar.svg

Ok, let's first discuss the situation w/o gauge fields

The Goldstone symmetry breaking mechanism says that the rotational symmetry is broken, i.e. that the state is not a V=0 but at V=Vo at some arbitrary angle θo. Now this angle θo breaks the rotational symmetry.

OK, the Goldstone bosons are nothing alese but field fluctutaions at constant V=Vo but fluctuating angle θ(x); b/c the potential is constant in θ-direction there is no force and therefore the fluctuations correspond to massless particles.

Now let's add gauge fields

That means that we do have a local symmetry where at every spacetime point x a local rotational symmetry in θ is present; that means that fluctuations in θ(x) do no longer correspond to particles but to gauge transformations and are therefore unphysical; any change in θ(x) at some spacetime point x can be rotated away by a gauge transformation. That's the reason why Goldstone bosons disappear.

Now back to Vo

http://en.wikipedia.org/wiki/Higgs_mechanism

The radial fluctuations in V still correspond to physical fluctuations, i.e. to Higgs bosons. But instead of having a fluctuation around V=0 we have one around V=Vo. This Vo corresponds to a mass term of the gauge fields (to understand that one has to look into the mathematical details).

The interesting thing is the following: massless gauge fields always correspond to two physical polarizations (two transversal polarizations for el.-mag. waves i.e. two d.o.f. for the photon), but massive vector fields have three degrees of freedom. So what happens is that one degree of freedom disappears from the Higgs field (the θ fluctuation) and reappers as the third (longitudinal) poolarization of the gauge field.

the gauge field eats the massless Goldstone boson
Thanks for the descriptions. I thought the Mexican Hat thing was only used as example in Inflaton fields where the idea is simplier but this can also be used in how the Higgs Eat the Goldstone too. I'd go back to the Mexican Hat portion in the book as the key to the understanding may lie there.

waterfall
#4
Feb26-12, 03:04 AM
P: 381
Why Eat Goldstone Boson?

Quote Quote by tom.stoer View Post
Have you ever seen the mexican hat potential?

http://en.wikipedia.org/wiki/Mexican..._hat_potential
http://en.wikipedia.org/wiki/File:Me...tial_polar.svg

Ok, let's first discuss the situation w/o gauge fields

The Goldstone symmetry breaking mechanism says that the rotational symmetry is broken, i.e. that the state is not a V=0 but at V=Vo at some arbitrary angle θo. Now this angle θo breaks the rotational symmetry.

OK, the Goldstone bosons are nothing alese but field fluctutaions at constant V=Vo but fluctuating angle θ(x); b/c the potential is constant in θ-direction there is no force and therefore the fluctuations correspond to massless particles.

Now let's add gauge fields

That means that we do have a local symmetry where at every spacetime point x a local rotational symmetry in θ is present; that means that fluctuations in θ(x) do no longer correspond to particles but to gauge transformations and are therefore unphysical; any change in θ(x) at some spacetime point x can be rotated away by a gauge transformation. That's the reason why Goldstone bosons disappear.

Now back to Vo

http://en.wikipedia.org/wiki/Higgs_mechanism

The radial fluctuations in V still correspond to physical fluctuations, i.e. to Higgs bosons. But instead of having a fluctuation around V=0 we have one around V=Vo. This Vo corresponds to a mass term of the gauge fields (to understand that one has to look into the mathematical details).

The interesting thing is the following: massless gauge fields always correspond to two physical polarizations (two transversal polarizations for el.-mag. waves i.e. two d.o.f. for the photon), but massive vector fields have three degrees of freedom. So what happens is that one degree of freedom disappears from the Higgs field (the θ fluctuation) and reappers as the third (longitudinal) poolarization of the gauge field.

the gauge field eats the massless Goldstone boson
Reviewing the book, I found it also described what you said. But I need some further information that not even wiki is clear about.

Are you saying in the gauge U(1) in electromagnetism for example, the gauge freedom in the phase from 0 to 360 degrees have corresponding Goldstone boson? Where are they? Aren't they detectable? How do you measure it if detectable and if not.. how do you make it detectable? Let's just take the example of electromagnetism and not electroweak to make distinct the concept. Thanks.
waterfall
#5
Feb26-12, 04:39 AM
P: 381
Quote Quote by waterfall View Post
Reviewing the book, I found it also described what you said. But I need some further information that not even wiki is clear about.

Are you saying in the gauge U(1) in electromagnetism for example, the gauge freedom in the phase from 0 to 360 degrees have corresponding Goldstone boson? Where are they? Aren't they detectable? How do you measure it if detectable and if not.. how do you make it detectable? Let's just take the example of electromagnetism and not electroweak to make distinct the concept. Thanks.
You may say that goldstone boson only come about in spontaneous symmetry breaking. But the em gauge applies to it.. or specifically wave functions or schroedinger equations with phases from 0 to 360 degrees. The symmetry occurs when the system is or particle is not yet emitted from for example the double slit emitter. After it it emitted. The phase is random from 0 to 360 degrees.. so here it is spontaneous symmetry breaking.. so where is the goldstone boson here?
tom.stoer
#6
Feb26-12, 09:38 AM
Sci Advisor
P: 5,364
Quote Quote by waterfall View Post
Are you saying in the gauge U(1) in electromagnetism for example, the gauge freedom in the phase from 0 to 360 degrees have corresponding Goldstone boson?
No.

A Goldstone boson is always related to the spontanerous brealdown of a global symmetry, not to a local gauge symmetry. So in U(1) gauge theory a Goldstone boson cannot exist.

In U(1) electromagnetism an abelian Higgs mechanism (w/o a Goldstone boson) could exist , but then you would have to add a charged scalar field with a Mexican hat potential; b/c we do not see such a field in nature this model is a purely theoretical contemplation. What we observe instead is the non-abelian Higgs mechanism (hopefully including the Higgs - to be discovered at the LHC) in the full el.-weak theory. This mechanism slightly more complicated but the basic idea is rather similar. In contrast to the abelian Higgs mechanism not the abelian but the non-abelian gauge symmetry is broken; that's why not the (abelian) photon but the (non.-abelian) W and Z are massive.
waterfall
#7
Feb26-12, 11:56 AM
P: 381
Quote Quote by tom.stoer View Post
No.

A Goldstone boson is always related to the spontanerous brealdown of a global symmetry, not to a local gauge symmetry. So in U(1) gauge theory a Goldstone boson cannot exist.

In U(1) electromagnetism an abelian Higgs mechanism (w/o a Goldstone boson) could exist , but then you would have to add a charged scalar field with a Mexican hat potential; b/c we do not see such a field in nature this model is a purely theoretical contemplation. What we observe instead is the non-abelian Higgs mechanism (hopefully including the Higgs - to be discovered at the LHC) in the full el.-weak theory. This mechanism slightly more complicated but the basic idea is rather similar. In contrast to the abelian Higgs mechanism not the abelian but the non-abelian gauge symmetry is broken; that's why not the (abelian) photon but the (non.-abelian) W and Z are massive.
You are trying to distinguish between U(1) gauge theory vs U(1) electromagnetism, are they not the same? Are you saying U(1) electromagnetism is a global gauge symmetry that is why you could induce an abelian higgs mechanism by adding a charged scalar field? In this case,
How do you add a charged scalar field with a Mexican hat potential to U(1) electromagntism and induce an abelian Higgs mechanism to give the photon mass? Thanks.
tom.stoer
#8
Feb26-12, 01:53 PM
Sci Advisor
P: 5,364
I am not trying to distinguish U(1) gauge theory and U(1) electromagnetism.

The abelian Higgs mechanism I am describing would be an extra ingredient which is not present in U(1) electromagnetism b/c there is no scalar charged particle.

Regarding the abelian Higgs mechanism please have a look here: http://en.wikipedia.org/wiki/Higgs_m...iggs_Mechanism
waterfall
#9
Feb26-12, 06:36 PM
P: 381
Quote Quote by tom.stoer View Post
I am not trying to distinguish U(1) gauge theory and U(1) electromagnetism.

The abelian Higgs mechanism I am describing would be an extra ingredient which is not present in U(1) electromagnetism b/c there is no scalar charged particle.

Regarding the abelian Higgs mechanism please have a look here: http://en.wikipedia.org/wiki/Higgs_m...iggs_Mechanism
I've been studying the site above and others. It ends with the statement:

"Furthermore, choosing a gauge where the phase of the vacuum is fixed, the potential energy for fluctuations of the vector field is nonzero. So in the abelian Higgs model, the gauge field acquires a mass. To calculate the magnitude of the mass, consider a constant value of the vector potential A in the x direction in the gauge where the condensate has constant phase. This is the same as a sinusoidally varying condensate in the gauge where the vector potential is zero. In the gauge where A is zero, the potential energy density in the condensate is the scalar gradient energy:"

But I can't get what is the whole purpose of the section when we know a photon has no mass and there is no charged scalar. Is it trying to say what would happen if there is a charged scalar? But it appears to be solving for someting definite and not at all concluding that this makes the photon massless because there is no charged scalar particle.
tom.stoer
#10
Feb27-12, 12:09 AM
Sci Advisor
P: 5,364
Quote Quote by waterfall View Post
But I can't get what is the whole purpose of the section when we know a photon has no mass and there is no charged scalar.
It's nothing else but a warm-up for the non-abelian Higgs mechanism in the full Glashow-Salam-Weinberg model with U(1) - SU(2) mixing due to the Weinberg angle and the spontaneous symmetry breaking in the weak sector.


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