# The homogeneous strength of the Higgs field

Mandragonia
Dear Physics Forum,

As I understand it, the Higgs field is a quantum field that stretches throughout our universe. Particles that carry mass (for example protons and electrons) acquire this property by interacting with the (local) Higgs field. I assume this interaction can be written in the form:

m(j;x,y,z,t) = alfa(j) * H(x,y,z,t)

Here alfa is a coupling constant and j denotes the particle type. H is the effective strength of the Higgs field and x,y,z,t are the space-time coordinates.

I find this intriguing! The mass of a particle used to be considered a fundamental property; just as charge, spin, parity (helicity). But now mass becomes an induced property that results from an interaction with an external field, and its coupling constant alfa(j) becomes the fundamental property.

There is an analogy to Classical Electromagnetism. Molecule can have a electron charge distribution that gives rise a permanent dipole moment d. Molecules can also have a charge distribution which is affected by an external electric field E. This gives to rise an induced dipole moment p = alfa * E. The constant alfa is the polarizability; a key-feature of the molecule.

The problem I have with the "induced mass" concept, is that it requires the Higgs field to be extremely homogeneous on different length and time scales. For example, if electrons in the Andromeda nebula have acquired a slightly different mass than in our galaxy, we would be able to detect this. The light from the Andromeda nebula would have slightly different spectral properties.

My questions are therefore on the homogeneity of the Higgs field.
1. The Higgs field interacts with matter. Hence -by symmetry- matter interacts with the Higgs field. To what extent can this interaction work as a source or sink, resulting in local fluctuations of the field strength?
2. Does the Higgs field have a property to dissipate or smooth out any fluctuations that occur in its local field strength? (e.g. like a conducting metal which responds efficiently and quickly to an excess charge that is applied to it, spreading it evenly out in a short period of time).
3. Is it possible that in certain regions of space (with a length scale L that may range from atomic to intergalactic) there is a small gradient term present in the Higgs field? Perhaps it is even possible that wave-like features can occur?

As I understand it, the Higgs field is a quantum field that stretches throughout our universe. Particles that carry mass (for example protons and electrons) acquire this property by interacting with the (local) Higgs field. I assume this interaction can be written in the form:

m(j;x,y,z,t) = alfa(j) * H(x,y,z,t)

Here alfa is a coupling constant and j denotes the particle type. H is the effective strength of the Higgs field and x,y,z,t are the space-time coordinates. I find this intriguing! The mass of a particle used to be considered a fundamental property; just as charge, spin, parity (helicity). But now mass becomes an induced property that results from an interaction with an external field, and its coupling constant alfa(j) becomes the fundamental property.
That's correct. The interaction is usually written me = v Ge, where v is the value of the Higgs field and Ge is the coupling constant. It turns out that v = 246 GeV. Giving mass to the fermions is important, of course, but the really fundamental role of the Higgs field is to break electroweak symmetry. Allowing the fermions to have mass comes as a bonus on top of that.

The problem I have with the "induced mass" concept, is that it requires the Higgs field to be extremely homogeneous on different length and time scales. For example, if electrons in the Andromeda nebula have acquired a slightly different mass than in our galaxy, we would be able to detect this. The light from the Andromeda nebula would have slightly different spectral properties.
The value of the Higgs field is just one of a couple dozen fundamental constants in the Standard Model. Another is the fine structure constant. Also the weak mixing angle, and so on. People have occasionally experimented with theories in which these constants are allowed to vary over cosmological distances, but as you point out it would have readily detectable effects, and no such effects have ever been seen.

dauto
Besides, the value v=246 GeV can be obtained as the value that minimizes the energy of the Higgs field. That is why the field in a vacuum has that value. Any other value would require a higher energy density and wouldn't be stable. By the way, v stands for "vacuum expectation value" Waves of higher energy density are possible just as waves on electromagnetic field are possible, but that would not be a vacuum. The photon is the particle associated with electromagnetic oscillations just as the Higgs particle is the particle associated with the Higgs field osculations.

Besides, the value v=246 GeV can be obtained as the value that minimizes the energy of the Higgs field.
It's obtained from the observed value of GF, the Fermi constant, the universal coupling constant that governs the rate of all weak interactions. The observed value is GF = 1.1 x 10-5 GeV-2. The relationship 1/(2v2) = G/√2 gives the value of v.

dauto
It's obtained from the observed value of GF, the Fermi constant, the universal coupling constant that governs the rate of all weak interactions. The observed value is GF = 1.1 x 10-5 GeV-2. The relationship 1/(2v2) = G/√2 gives the value of v.

Yes that's also true and it shows that speculating on the possibility that the value of v might be different elsewhere in the universe is equivalent to speculating on the possibility that the electric charge might be different elsewhere in the universe. Possible, but definitely not likely.

Dear Physics Forum,
3. Is it possible that in certain regions of space (with a length scale L that may range from atomic to intergalactic) there is a small gradient term present in the Higgs field? Perhaps it is even possible that wave-like features can occur?

When you read an announcement that the Higgs Boson has been discovered at the LHC, what they are seeing is wave-like oscillations of the Higgs field around its background value.

When you read an announcement that the Higgs Boson has been discovered at the LHC, what they are seeing is wave-like oscillations of the Higgs field around its background value.
Only if you believe that all the elementary particles we observe are wave-like oscillations of the vacuum. The Higgs field is not a classical field, and does not support "wave motion." It's a quantum field, and the Higgs boson is its quantum excitation.

dauto
Only if you believe that all the elementary particles we observe are wave-like oscillations of the vacuum. The Higgs field is not a classical field, and does not support "wave motion." It's a quantum field, and the Higgs boson is its quantum excitation.

In what sense do you mean it doesn't support wave motion? Ii obeys the Klein and Gordon wave equation, after all...

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Only if you believe that all the elementary particles we observe are wave-like oscillations of the vacuum. The Higgs field is not a classical field, and does not support "wave motion." It's a quantum field, and the Higgs boson is its quantum excitation.

Hmm. I'm not sure I see where we are disagreeing. If I said that a photon was "a wave-like oscillation of the electromagnetic field", would you disagree? Perhaps you would. How would you describe it then? As a "quantum excitation of the electromagnetic field"? Are you taking issue with the phrase "wave-like oscillation"? Photons are clearly "wave-like" wouldn't you agree? If so, do you think the Higgs boson is not "wave-like"?

Hmm. I'm not sure I see where we are disagreeing. If I said that a photon was "a wave-like oscillation of the electromagnetic field", would you disagree? Perhaps you would. How would you describe it then? As a "quantum excitation of the electromagnetic field"? Are you taking issue with the phrase "wave-like oscillation"? Photons are clearly "wave-like" wouldn't you agree? If so, do you think the Higgs boson is not "wave-like"?
How about other particles. Would you consider a gluon a "ripple in the gluon field"? Is a W boson a "wave motion"? It's the classical context that I'm objecting to. Photons and gravitons can form classical waves. Other particles including the Higgs cannot, especially considering the lifetime of the Higgs is 10-26 sec.

I've seen this explanation given in some popularized accounts - that the Higgs boson is a "ripple in the Higgs field," and while you and I know how to reinterpret what they're saying, I think it's a misleading analogy for nonphysicists. It makes questions like we see in the OP sound perfectly reasonable, in which the Higgs field is not just a background constant value, rather it behaves like an ocean that can be expected to vary ("gradients") in large amplitude waves over space and time.

How about other particles. Would you consider a gluon a "ripple in the gluon field"? Is a W boson a "wave motion"? It's the classical context that I'm objecting to. Photons and gravitons can form classical waves. Other particles including the Higgs cannot, especially considering the lifetime of the Higgs is 10-26 sec.

I've seen this explanation given in some popularized accounts - that the Higgs boson is a "ripple in the Higgs field," and while you and I know how to reinterpret what they're saying, I think it's a misleading analogy for nonphysicists. It makes questions like we see in the OP sound perfectly reasonable, in which the Higgs field is not just a background constant value, rather it behaves like an ocean that can be expected to vary ("gradients") in large amplitude waves over space and time.

I see your point, especially given the extremely short lifetime. If the Higgs boson were long-lived, since it's a boson, you could imagine a large number of them in the same quantum state, and then you would have a classical wave. But with the short lifetime that can't really happen. OK then, I'm happy with calling the Higgs boson "a quantum excitation of the Higgs field."

Mandragonia
Photons and gravitons can form classical waves. Other particles including the Higgs cannot, especially considering the lifetime of the Higgs is 10-26 sec.

I acknowledge that the very short lifetime of the Higgs boson is an important factor.

However, it would seem to me that -perhaps- this is balanced by the requirement of a very high rate of production of these virtual particles per unit of volume (cubic centimeter) and per unit of time (second). Otherwise there would not be a sufficient number of successful interactions per unit of time with a test particle with mass. As a result the acquired mass of the test particle would not be perceived as a perfectly constant value, but instead jittery due to statistical fluctuations.

Mentor
Other particles including the Higgs cannot, especially considering the lifetime of the Higgs is 10-26 sec.
1.56*10-22, assuming a standard model Higgs boson at 126 GeV.
Still too short to form classical waves, of course.

Mandragonia said:
this is balanced by the requirement of a very high rate of production of these virtual particles per unit of volume (cubic centimeter) and per unit of time (second).
There is no "rate of produced virtual particles". That does not exist.

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However, it would seem to me that -perhaps- this is balanced by the requirement of a very high rate of production of these virtual particles per unit of volume (cubic centimeter) and per unit of time (second). Otherwise there would not be a sufficient number of successful interactions per unit of time with a test particle with mass. As a result the acquired mass of the test particle would not be perceived as a perfectly constant value, but instead jittery due to statistical fluctuations.
The fermion masses are a result of interaction with the constant value of the background Higgs field. The short lifetime applies to the creation and subsequent decay of the Higgs boson, which is not involved in this.

Mandragonia
There is an explanation of the "acquired mass" phenomenon on Wikipedia. For a system of fermions interacting with a scalar field one can construct the Lagrangian, with an interaction term assumed to be of of the Yukawa type. Under certain conditions one may split off a constant (background) term from the scalar field. The corresponding Yukawa term simplifies further, so that it behaves exactly as the Dirac mass term of the fermion field.

Fine! What surprises me though, is that the fermion masses acquired by this process are considered to be the real masses of the particles..... In other branches of Physics one restructures equations in similar ways, often with the result that the initial parameter recombine into new ones. For example in Condensed Matter Physics the interaction of electrons with the lattice gives rise to an apparent extra mass term. The net result of the calculation is called the effective mass of the electron.

Mentor
It is a mass those particles have everywhere - electrons can leave a metal (or get in a different band), but they can never leave the Higgs field.

Mandragonia
In Condensed Matter Physics the effective mass is a useful concept for understanding the electron's response to forces (e.g. an externally applied electric field). Essentially, this is done by redefining the electron's inertial mass. The effective mass can be ~100 times larger than its real mass.

On the other hand, if one considers electron-positron annihilation in solids, the energy of the 2 gammaray photons that are emitted varies only little from the standard value of 0.511 MeV.

This demonstrates that the effective mass concept is restricted to inertial mass applications (kinetic energy, momentum, mobility). It does not extend to other electron properties, such as its rest mass.

On the other hand, if one considers electron-positron annihilation in solids, the energy of the 2 gammaray photons that are emitted varies only little from the standard value of 0.511 MeV. This demonstrates that the effective mass concept is restricted to inertial mass applications (kinetic energy, momentum, mobility). It does not extend to other electron properties, such as its rest mass.
Thanks, this is a good argument to counter the claim that the Higgs phenomenon is analogous to the effective mass in Condensed Matter Physics.

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Mandragonia
Perhaps the Higgs phenomenon is more profound, because it is based on relativistic quantum physics. The Dirac equation features a mass term, which serves both as rest mass and inertial mass. In fact, there is no distinction between the two. Whereas in Condensed Matter Physics one solves the non-relativistic Schroedinger equation. It describes electron wave functions using their inertial mass as parameters; there is no room for effects involving the rest mass of the electron.

Mandragonia
I have another question.

The Higgs phenomenon finds its origin in relativistic quantum effects. Yet the resulting expression for the mass-giving interaction [m = cc*H(0)] appears to be at odds with this. The interaction is not quantized and may not be quantizable (there is no Feynman diagram for an electron sensing the Higgs and thus getting mass). Furthermore the effect appears to be instanteneous, instead of occurring at the light speed c.

Mentor
(there is no Feynman diagram for an electron sensing the Higgs and thus getting mass)
Sure there is.
I don't think the rest makes sense as a question.

dauto
The effect is instantaneous because the field is everywhere.

Mandragonia
Sure there is.

I have seen several Feynman diagrams involving the Higgs boson, but none that describe the interaction with the Higgs background field.

Mentor
There is a picture that symbolizes the interaction with the field here (German): But you can just draw a graph of an electron, emitting a Higgs and absorbing it back again.

Mandragonia
The effect is instantaneous because the field is everywhere.

Exactly. By construction, the overall effect of the Higgs interaction is static. Which is a bit unfortunate, as it provides no insight into the underlying (relativistic quantum) physics of the interaction.

Mandragonia
There is a picture that symbolizes the interaction with the field. But you can just draw a graph of an electron, emitting a Higgs and absorbing it back again.

Thank you very much! Could you please explain to me what you mean with "emitting a Higgs"? It sounds to me like a virtual particle associated with the Higgs background field, but (as I have been informed on this forum) this is not the Higgs boson.

But you can just draw a graph of an electron, emitting a Higgs and absorbing it back again.
No, that graph would be part of the electron self-energy, not the same thing.

Jedi_Sawyer
I think the concept of the Higgs field giving mass to particles and the concept of gravitons are related somehow. So is a Higgs field carrier the graviton? Maybe once the Higgs Particle turns into a field with carriers, it is automatic that the Higgs field will be related to the mass it is giving a particle and that is why it is homogenous.

There is a picture that symbolizes the interaction with the field here (German): That diagram is sorta ok, but it would be better if one drew an x on the insertions, to emphasize the fact that it's really an interaction with the Higgs condensate that is allowing the helicities to flip back and forth and thus providing mass to the fermions.

I think the concept of the Higgs field giving mass to particles and the concept of gravitons are related somehow. So is a Higgs field carrier the graviton? Maybe once the Higgs Particle turns into a field with carriers, it is automatic that the Higgs field will be related to the mass it is giving a particle and that is why it is homogenous.

Absolutely not, there is no relationship between the Higgs field and gravitons.

Mentor
Thank you very much! Could you please explain to me what you mean with "emitting a Higgs"? It sounds to me like a virtual particle associated with the Higgs background field, but (as I have been informed on this forum) this is not the Higgs boson.
The real Higgs boson is an excitation of that field, the virtual ones are a way to visualize perturbation theory.

No, that graph would be part of the electron self-energy, not the same thing.
Still a part of its mass, and don't forget the other part of my post.

kurros
Still a part of its mass, and don't forget the other part of my post.

But before symmetry breaking there are still higgs loop contributions to the fermion propagators, yet they don't get mass from these. This is actually a question: is the reason for this that the gauge symmetry protects the fermion masses from these corrections? Until we break the gauge symmetry of course?

Perhaps related: photons have no direct interaction with the Higgs, but they do at loop level, and so the photon propagator has Higgs bosons floating around in it. Nevertheless it remains massless.

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dauto
But before symmetry breaking there are still higgs loop contributions to the fermion propagators, yet they don't get mass from these. This is actually a question: is the reason for this that the gauge symmetry protects the fermion masses from these corrections? Until we break the gauge symmetry of course?

Perhaps related: photons have no direct interaction with the Higgs, but they do at loop level, and so the photon propagator has Higgs bosons floating around in it. Nevertheless it remains massless.

Well, the problem is that the mass term connects the left- and right- handed spinors and those spinors belong to two different representations of the gauge group symmetry so there is no way to create a mass term that is a singlet (invariant) under the symmetry. So a explicit mass term in the Lagrangean would indeed break the gauge symmetry. The way around that is to create an interaction of those two fields with a third field. That way it is possible to create a term in the lagrangean that involves both the left- and the right-handed spinors which is also a gauge invariant. That will work if that third field is a complex doublet under the weak isospin gauge group. That third field is, of course, the Higgs field. When the Higgs field acquires a non-vanishing vacuum expectation value, its interaction term with the fermions turn into mass terms.

A mass for the photon would break electromagnetic gauge invariance which survives the symmetry breaking, so it must remain massless.

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kurros
Hmm, yeah I get all that, but I guess my question is more about why such mass terms are not generated during renormalisation of the fields. The fields that have tree-level masses receive corrections to their masses as mfb describes, but it seems that if no tree-level mass exists then one doesn't get generated by such corrections. My postulate was that this is because the original gauge symmetry somehow arranges for all such corrections to cancel each other out or some such; I guess this is related to anomalies or something right? The gauge symmetry is preserved at the quantum level.

dauto
Hmm, yeah I get all that, but I guess my question is more about why such mass terms are not generated during renormalisation of the fields. The fields that have tree-level masses receive corrections to their masses as mfb describes, but it seems that if no tree-level mass exists then one doesn't get generated by such corrections. My postulate was that this is because the original gauge symmetry somehow arranges for all such corrections to cancel each other out or some such; I guess this is related to anomalies or something right? The gauge symmetry is preserved at the quantum level.

Yes to everything you said, including the part where anomalies must cancel out otherwise the symmetry is not preserved by quantum corrections. Assuming that the anomalies do cancel, radiative corrections to the photon mass mass vanish by gauge symmetry. Note though that in general it is possible to build theories where a particle mass is created by radiative effects even if that mass was absent from the tree-level (bare particles) Lagrangian.