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Electric Charge vs Mass in Gauge Bosons

  1. Jan 10, 2015 #1
    Is there any significance to the fact that:
    • The electromagnetic and strong interactions have gauge bosons with no electric charge that are massless; and
    • The weak interaction has two massive gauge bosons which do have electric charge.
    If there is a significance to this 'observation' then where does the Z0 gauge boson fit in?


    IH
     
  2. jcsd
  3. Jan 10, 2015 #2

    Doug Huffman

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  4. Jan 10, 2015 #3
    Do you mean why would *weak* gauge bosons have *electric* charge? Sure, there is a good reason; it is because the weak and electromagnetic interaction are unified in electroweak theory.

    In pure unbroken gauge theories, all the gauge bosons are only charged under whatever gauge group it is which generates them (except for U(1) gauge bosons), and they are all massless. So gluons have no weak or electric charge because QCD is essentially completely separate to electroweak theory. The same logic doesn't apply to the electroweak sector because of electroweak symmetry breaking, and it is the particular way in which it is broken which leaves you with some neutral and some charged electroweak gauge bosons, and also generates masses for some of them.
     
  5. Jan 10, 2015 #4

    samalkhaiat

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    It simply means that the [itex]U_{em}(1)[/itex] and [itex]SU_{c}(3)[/itex] gauge fields are REAL fields.

    Weak interaction is the result of gauging the global [itex]SU(2)\times U(1)[/itex] symmetry. The GAUGE group [itex]SU(2)[/itex] has 3 REAL MASSLESS gauge FIELDS [itex]W^{ i }_{ \mu }[/itex], [itex]i = 1, 2 , 3[/itex]. Therefore, they have no em charge. The corresponding MASIVE and CHARGED gauge BOSONS (don't confuse them with gauge fields) of the weak interaction are the following combinations
    [tex]W^{\pm}_{ \mu } = \frac{ 1 }{ \sqrt{ 2 } } ( W^{1}_{ \mu } \mp i W^{ 2 }_{ \mu } )[/tex]
    These are charged because they are complex fields, i.e. they admit a GLOBAL [itex]U(1)[/itex] symmetry. The 3rd gauge BOSON has no charge because it is given by the follwing REAL combination
    [tex]Z^{0}_{ \mu } = W^{ 3 }_{ \mu } \sin \theta - B_{ \mu } \cos \theta ,[/tex]
    where [itex]B_{ \mu }[/itex] is a MASSLESS REAL [itex]U(1)[/itex] gauge FIELD.

    Sam
     
    Last edited: Jan 10, 2015
  6. Jan 10, 2015 #5

    samalkhaiat

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    No, there are 3 of them. There are, always, as many gague fields as there are generators of the gauge group.
    This make them 4?
    Which chiral symmetry?
     
  7. Jan 10, 2015 #6

    ChrisVer

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    :s How is reality connected with charge?
    Because of the complex conjugation?
     
  8. Jan 10, 2015 #7
    I don't understand the question...
     
  9. Jan 10, 2015 #8

    ChrisVer

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    Is it a "must" for a real field to be chargeless? For scalars that's obvious....
     
  10. Jan 10, 2015 #9

    samalkhaiat

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    Yes. The electric charge IS the Noether number associated with the GLOBAL [itex]U(1)[/itex] symmetry. This vanishes for REAL FIELDS.
     
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