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Higgs field

  1. Nov 13, 2008 #1
    Why is the Higgs field a scalar field? I understand if it is one, it will have no spin and no angular momentum. But understanding that a particle is a scalar seems to me a leap of faith. What am I not getting?
  2. jcsd
  3. Nov 13, 2008 #2
    I apologize to the readers. I should have read farther down the list to see some other posts on this subject. I gathered that somewhere within the symmetry a breaking happens. Is the breaking the responsible property for the scalar characteristic? I have a feeling I'm confusing terms which is confusing readers. This stuff is hard to absorb.
  4. Nov 13, 2008 #3
    The Higgs field arises out of the spontaneous symmetry breaking during inflation. I don't know the precise mathematics formalism, but Higgs, strong, weak, electromagnetic fields pop out of the higher energy unified force...gravity too, but I think that is separate....
  5. Nov 13, 2008 #4
    Let me explain it in my own words what I think you said. Symmetry happens with every fresh region of the universe being created as it expands. In the expansion, Higgs field is the first to break. In some ways, I picture the Higgs field like the curtain covering the grainy constituents ready to enter the energy barrier before entering the observable universe. In that respect, the Higgs field offers 'skin' to the little parts. However, my original question was how can a particle field resemble a scalar field? I'm thinking that it is a composite field which has the appearance of a magnitude, because it is a ratio of fields. Am I in left field with this idea?
  6. Nov 13, 2008 #5
    Ok, imagine this.

    You have a simple pendulum and you let is swing back and forth. (I'll do this with a John Baez-esque ASCII illustration):


    Where is the average position of the pendulum bob? You should be able to convince yourself that it is zero. It lives just as much to the right as it does to the left. So the whole system is symmetric under the exchange of left-right.

    Now take the whole pendulum and tilt it. Now let it swing back and forth. (The dots are placeholders!)


    Now, the average position is no longer zero---you've broken the symmetry by tilting the pendulum.

    The same thing happens with the higgs field---it's expected value is no longer at zero. Technically, we say that the higgs gets a vacuum expectation value. This is just a fancy way to say that the vacuum isn't symmetric when the higgs gets a vev.

    Ok. So what happens when the higgs gets a vev? Well, all of the symmetries that the higgs was charged under are no longer symmetries. I can't think of a non-technical way to explain this, so apologies :) Take it on faith, for now, that this is true.

    The final piece of the puzzle is that the higgs MUST be a scalar. We know that GR is a good theory for low energy physics---we can calculate things with it, and we've tested it to a fantastic accuracy. GR depends on a certain symmetry called SO(3,1). This is the symmetry of rotations and boosts, and says (roughly) that if you rotate your coordinate frame, you should get the same physics.

    So we know we have a broken symmetry (called the electroweak symmetry). (We know this because the electroweak symmetry is a short range force.) But we also know that GR (gravity) is NOT broken. This means that the higgs (which does the symmetry breaking) must be charged under the electroweak symmetry, but it can't break the SO(3,1) symmetry of GR. The only type of particle that does this is a scalar particle. More succinctly, it is CALLED a scalar particle because it doesn't have any transformation properties under the SO(3,1) of gravity. Another way to say it is that the spin of the particle is zero---the spin relates to the way that the particle is acted on by the symmetries of GR.

    Hopefully this wasn't too confusing :)
  7. Nov 14, 2008 #6
    Am I understanding this right that the charged-ness, the turned on position, of the Higgs field is a consequence of the symmetry? Or is it the breaking? In other words, when is the charged under property activated? When I hear talks about scalar fields, I envision something like a sea of speeds without any objects being sped along. If this analogue is correct, then what type of vibration would a Higgs represent? Scalar fields from what I've read are indicators, not the occurrence in question, such as temperature distribution. Rather than trying to beleaguer the point of why or how a scalar particle will raise a field beyond the breaking point, I've accepted it as true to hurry this topic. However, I can't see the relationship of what such a field means in physical constructs. Is it a wave or matter or communication within the elements of gravitational bodies?
  8. Nov 14, 2008 #7
    Well, the analogy above has to be taken with a grain of salt. "Position" in field space corresponds to a value that the field takes on. The fact that the higgs takes on a value everywhere (the same value) is a consequence of the symmetry breaking.

    Yeah, this is roughly right. The higgs field takes a constant value everywhere in space. Particles move through this field and interact with it, like a fish moves through water. The local value of the field (which is, remember the same everywhere) tells the particle how much it weighs.

    There is another analogy, with superconducters---I've never particularly digested it (i.e. I don't understand it), but perhaps you can get something out of it: http://en.wikipedia.org/wiki/Higgs_mechanism#Superconductivity
  9. Nov 15, 2008 #8


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