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Question on Z,W,photon fields

  1. Jul 11, 2012 #1
    I was looking at the Lagrangian density:

    L= [itex]-\frac{1}{4}[/itex] [itex]Α^{α}_{μν}[/itex][itex]Α^{μν}_{α}[/itex][itex]-\frac{1}{4}[/itex] [itex]Β^{}_{μν}[/itex][itex]Β^{μν}_{}[/itex][itex]+\bar{e}_{R}[/itex](i∂-g'B)[itex]e_{R}[/itex]
    +[itex]\bar{L}_{e} (i\partial-\frac{g'}{2}B+\frac{g}{2}τ^{α}Α_{α})L_{e}-G_{e}\frac{f+p(x)}{\sqrt{2}}(\bar{e}_{R}e_{L}+{e}_{R} \bar{e}_{L})[/itex]
    [itex]+\frac{1}{2}({\partial}_{μ} p(x))^{2} +\frac{1}{8} (f+p)^{2} [ (g' {B}_μ-g A_{μ}^{3})^{2} + g^{2} ({Α}_{μ}^{1} Α^{μ1} +{Α}_{μ}^{2} Α^{μ2}) ] - V( \frac{(f+p(x))^{2}}{2})[/itex]

    Now I will try to explain what everything is (although i don't know if it has anything to do with the question I want to make):
    Le is the doublet of Isospin
    [itex]{L}_{e}=( {v}_{e} , {e}_{L} ) [/itex]
    [itex]{e}_{L}=\frac{1}{2} (1-γ^{5}) {Ψ}_{e} [/itex] the left component of the electron spinor-wavefunction
    with hypercharge -1

    [itex]{e}_{R}=\frac{1}{2} (1+γ^{5}) {Ψ}_{e} [/itex] the right component.
    with hypercharge -2

    there are the 3 gauge fields [itex] {A}_{μ}^{α}, α=1,2,3 [/itex] for the SU(2) with coupling constant g.
    the 1 gauge field [itex] {Β}_{μ} [/itex] for Uy(1) with coupling constant g'/2

    and for the symmetry breaking I used 4 scalar fields (2 complex)
    Φ=(Φ* Φ0)
    with hypercharge +1
    and the well known Potential
    [itex] V(Φ*Φ)= {μ}^{2}(Φ*Φ)+λ{Φ*Φ)^{2} [/itex]

    One thing more about my notation, everywhere you see A,B or the partial derivative, it is DASHED. I don't know how to make it...

    then to get the above lagrangian I expanded my initial lagrangian over the field Φ around the value:
    Φ(x)= ( 0 [f+p(x)]/√2 ) where f2= -μ2/λ >0

    then you get the above baby monster....


    Now my question
    When you define the new fields Z,W,A:
    [itex]W^{\pm}_{μ}=\frac{1}{\sqrt{2}}(Α^{1}_{μ}\mp i Α^{1}_{μ})[/itex]
    [itex]Ζ_{μ}=(g^{2}+g'^{2})^{-\frac{1}{2}}(-gΑ^{3}_{μ}+ g' B_{μ})[/itex]
    [itex]A'_{μ}=(g^{2}+g'^{2})^{-\frac{1}{2}}(g B_{μ}+ g' A^{3}_{μ})[/itex]

    I see that W is written in respect to the initially defined gauge field of the SU(2) so it means it could couple the isospin doublets (i hope my term was correct) but it cannot couple charges (because it does not have the Uy(1) gauge field B in it). Something that is wrong because W is a charged particle and it COULD interact electromagneticallly...

    On the other hand, Z seems to be able to interact electromagnetically because it has the B field...

    And finally the Photon A' can couple the isospin doublets? Coz it has the 3rd field A3μ

    What am I thinking wrong of it about?
     
  2. jcsd
  3. Jul 11, 2012 #2
    If you write out the gauge kinetic term for the original SU(2) field in terms of the W, Z, and A' fields, I think you will find that there are terms that couple the W to the A': I believe there will be a WWA' term and a WWA'A' term. The A' field is the photon field; thus the W couples to photons, i.e. it is charged under electromagnetism. However, you shouldn't find any ZZA' or ZZA'A' coupling. Thus the Z doesn't couple to the photon; i.e., it is electrically neutral.

    I'm not sure what you mean here:

    If you want to see how the A' couples to matter, write out the original term coupling the gauge fields to matter in terms of the W, Z, and A' fields. You should be able to obtain a formula for the electrical charge of a particle in terms of its hypercharge (charge under U(1), usually written Y) and its weak isospin (its value for the third component of SU(2), usually written T_3).
     
  4. Jul 11, 2012 #3
    what I meant with the question was that I wrote the field of the photon A' as Bμ and Aμ sum...
    Some lines above I determined B,A fields as the one working on U(1) hypercharge and SU(2) isospin respectively.

    So I meant exactly the same thing I meant for Ws and Zs. That for the photon field A' I have contribution from the SU(2) isospin field A3μ which couples the isospin doublets.... this seemed strange....
     
  5. Jul 11, 2012 #4
    Yes, since the photon field is in part composed of A^3 the photon will couple to all weak isospin doublets. And since this part of the coupling is proportional to T_3, the coupling of the photon to the "up" component of the doublet will differ from the coupling to the "down" component of the doublet. This is why e.g. the up quark and down quark have an electric charge that differ by one, and similarly the electron and the electron neutrino have a charge that differ by one.

    The idea in electroweak symmetry breaking is that SU(2)_isospin x U(1)_hypercharge breaks down to U(1)_electromagnetism, but U(1)_electromagnetism is not the same as U(1)_hypercharge. As you've seen, U(1)_electromagnetism is generated by a linear combination of the hypercharge generator and the third isospin generator. Thus the photon couples to anything that is charged under either SU(2)_isospin or U(1)_hypercharge [unless these couplings end up cancelling out, as with the neutrino].
     
  6. Jul 11, 2012 #5
    ^_^ Thank you for the help, I think I get it now...
     
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