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Help understanding the Higgs field?

  1. Jul 26, 2012 #1
    So the way I understand it is that certain particles move through the Higgs field and encounter no resistance, giving it no mass. The others that do encounter resistance are the ones that have mass. But if increasing resistance means increasing mass, why wouldn't things become infinitely massive? Also, why don't all of the Higgs bosons come together, seeing as they basically are mass, shouldn't they just keep attracting each other through gravity? Lastly, (for now at least) how can a W or Z boson be less massive than a single Higgs boson? :confused: :confused: :confused:
  2. jcsd
  3. Jul 26, 2012 #2
    The 'resistance' picture is very misleading. What really goes on is quite different - we say the Higgs field is coupled to the fermionic fields that represent the matter particles. In empty space, the Higgs field takes a value - called the vacuum expectation value (or VEV) - at every point. When particles interact, we can represent this by a Feynman diagram (do a google search for some pictures of what a Feynman diagram looks like). The interactions occur at points called vertices. At each vertex, the particle interacts with the VEV of the Higgs field. This is called a Yukawa interaction, and we plug it into the Standard Model Lagrangian (a Lagrangian is a term that essentially express the energy of the system, in the form kinetic minus potential energy) and declare that the fermions have mass given by the value of the VEV, and the strength of the coupling of each fermionic field to the Higgs field.

    Gravity is negligible at the scale of particle physics. Only at extremely high energies is gravity relevant in particle physics.

    Why does is seem odd that the W and Z bosons have less mass than a Higgs boson?
  4. Jul 26, 2012 #3
    Wait, so just to make sure I'm understanding, fields describing fermions are coupled to another field, the Higgs field, and so they interact. The Higgs field comes with a value that, since it comes with the fermionic field passes that value onto the fermion?

    And if the Higgs takes a value at every single point in space, wouldn't any small amount of gravity be significant? Do they not interact with each other for some reason?

    Also, is it the same for leptons?

    Many thanks
  5. Jul 26, 2012 #4
    Oh and it seemed odd that they had less mass because I was thinking of the Higgs as like a single "unit" of mass, not as a particle whose interactions constitute mass <--(correct?)
  6. Jul 26, 2012 #5
    Essentially, that's the idea. The Higgs field couples to fermions, and the strength of the coupling determines the mass of those fermions.
    Well, yes, the Higgs field carries a potential. But this is irrelevant. Similar to how gravity is irrelevant when describing electromagnetism and the strong force.
    Leptons are fermions, so yes.
  7. Jul 26, 2012 #6
    Ahhh thanks!
  8. Jul 26, 2012 #7
    The Higgs boson is the quantum of the Higgs field. It doesn't make up a unit of mass, since it's the VEV of the Higgs field that determines fermion masses.
  9. Aug 4, 2012 #8
    So since the fields are coupled the VEV will always have/be giving a value because an interaction is always occuring? Is that how it works?

    Sorry...I'm trying to think of a way to visualize this in my mind and it's proving to be pretty difficult..
  10. Aug 4, 2012 #9
    Right. Every fermionic field has a coupling constant that determines how strongly it couples to the Higgs field. Together with the Higgs VEV, this determines the particle masses.

    Don't worry, there really isn't anything you can do to visualize it. That's why I said it's misleading to describe it as some kind of 'molasses'.
  11. Aug 13, 2012 #10
    Ahhhh thanks so much for all of your help!!
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