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Relationship between higgs field and EM field

  1. Dec 16, 2011 #1
    So I'm trying to understand how the Higgs field and Higgs boson are different from the EM field and the photon. Firstly how I understand the photon is that It is the quanta of energy that an electromagnetic field oscillating at a certain frequency can give to some other system. The photon is the smallest possible ripple(of a certain frequency) in the EM field.

    I'm inclined to think that the same relationship holds for the Higgs field and the Higgs boson. The way I understand it is that It requires the amount of energy that they are using at the LHC to create the 'ripple' in the Higgs field. The Higgs field exists at a constant 'value' everywhere.

    I'm just learning about this stuff myself so I apologize for my lack of knowledge about things that are probably vital to the understanding of the subject. I'm a first year grad student I haven't had any formal training in particle physics though. I would love to hear corrections to my thinking.
     
  2. jcsd
  3. Dec 16, 2011 #2

    Bill_K

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    Sounds good to me!
     
  4. Dec 17, 2011 #3
    Well so can you think of the higgs field like an em field except that it has a constant value everywhere whereas the em field can be zero? Obviously its different in the way it interacts with things. However if it is like an em field with a constant value then it has to have a preferred direction right? That seems to imply that it can't be a spatial vector and if it has a constant value everywhere that just leaves the analogue of a scalar field which I'm sure is too simple...
     
  5. Dec 18, 2011 #4
    It is quite simple indeed, but you get it right again. That is, after symmetry breaking, there is only one scalar Higgs field remaining, and in the vacuum it is a non-zero constant, at the classical level (the Higgs particle is the deviation from this constant, in the potential direction). But hey, fermions have a constant mass, so they'd better couple to a constant field !

    The story is a little less trivial for giving mass to the weak vector bosons.
     
  6. Dec 18, 2011 #5
    Wow alright that makes sense, thank you. So the only way that this field can deviate from its constant value is with a higgs boson, which we are looking for.

    So fermions interact with the field more simply than the vector bosons which I assume means the W and Z bosons. I've heard the analogy that mass is the drag these particles feel as they move through the higgs field. Can anyone elaborate on any substance to that analogy? It seems bogus because you would have to have a velocity to have mass. Or maybe they are talking about an "acceleration" drag.
     
  7. Dec 18, 2011 #6
    The analogy was made popular by a cartoon for a context in the UK long ago I believe. But I also think there is some value in the picture, even though it does not provide any hint as to the "mexican potential" or the breaking itself in general.

    One may think at first as all particles being massless. Fermions comes in two independent copies except for opposite helicities, which we count as two different fundamental particles. They propagate freely without mixing at the speed of light. Once the Higgs field is introduced and a gauge is picked, it acquires spontaneously a vacuum expectation non-zero value. This generates a coupling term between the two helicity species, which is the mass term we were missing. In the old days we called this picture "zitterbewegung" if you want to look it up. The "zig-zag" motion at the speed of light generate an effective velocity lower for larger coupling between the two helicities (because the zig-zag transition occurs more rapidly).
     
  8. Dec 19, 2011 #7
    Hmm interesting. Any recommendation for a introductory book to learn what you're talking about in more detail? I'm a first year grad student.

    Thanks a lot for your responses.
     
  9. Dec 20, 2011 #8
    I'll second this request on quality reading material.
     
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