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Vector Boson Propagator and gauge

  1. Sep 1, 2014 #1


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    Well I'm trying to understand the difference between these propagators:

    [itex] \frac{g_{\mu \nu}}{k^{2} - m^2 + i \epsilon}[/itex]


    [itex] \frac{g_{\mu \nu}+ \frac{ k_{\mu} k_{\nu}}{m^{2}}}{k^{2} - m^2 + i \epsilon}[/itex]

    My professor told me that they are different gauges, and the from the second you rule out the goldstone modes. Can someone explain it to me a little better or refer me to some textbook? I don't also get the meaning of goldstone modes- if you have a massive vector boson then you don't have goldstone modes [they become the longitudial dofs]

    Also I am having one more question. If the first doesn't kill out the goldstone bosons, then can someone -after arriving at the end result- kill them? Maybe by a gauge transformation? and what gauge transformation?
    Last edited: Sep 1, 2014
  2. jcsd
  3. Sep 1, 2014 #2


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    I am not sure what your prof meant by "killing off the Goldstone modes". The two expressions correspond to different gauges. If you put a parameter ##\xi## in front of the ##k^\mu k^\nu## term, upon summing up any gauge invariant combination of Feynman diagrams (in particular the sum of the diagrams of a given loop order), the ##\xi## dependent terms will cancel out, showing that adding that term does not change anything to the final result. I guess it is possible to think of these extra terms as being associate to the Goldstone modes (like some pieces of the ##g^{\mu \nu}## piece in QED correspond to the longitudinal modes) .
  4. Sep 3, 2014 #3


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    You need to look at a QFT text for details; try Srednicki.
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