The homogeneous strength of the Higgs field

  1. 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?

    I thank you in advance!
  2. jcsd
  3. Bill_K

    Bill_K 4,157
    Science Advisor

    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 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.
  4. 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.
  5. Bill_K

    Bill_K 4,157
    Science Advisor

    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.
  6. 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.
  7. phyzguy

    phyzguy 2,489
    Science Advisor

    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.
  8. Bill_K

    Bill_K 4,157
    Science Advisor

    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.
  9. In what sense do you mean it doesn't support wave motion? Ii obeys the Klein and Gordon wave equation, after all...
    Last edited: Oct 15, 2013
  10. phyzguy

    phyzguy 2,489
    Science Advisor

    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"?
  11. Bill_K

    Bill_K 4,157
    Science Advisor

    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.
  12. phyzguy

    phyzguy 2,489
    Science Advisor

    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."
  13. 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.
  14. mfb

    Staff: Mentor

    1.56*10-22, assuming a standard model Higgs boson at 126 GeV.
    Still too short to form classical waves, of course.

    There is no "rate of produced virtual particles". That does not exist.
    Last edited: Oct 15, 2013
  15. Bill_K

    Bill_K 4,157
    Science Advisor

    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.
  16. 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.
  17. mfb

    Staff: 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.
  18. 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.
  19. Bill_K

    Bill_K 4,157
    Science Advisor

    Thanks, this is a good argument to counter the claim that the Higgs phenomenon is analogous to the effective mass in Condensed Matter Physics.
    Last edited: Nov 5, 2013
  20. 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.
  21. 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.
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