Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Polarization of gauge bosons and gauge choice

  1. Mar 15, 2017 #1
    Consider the following facts:

    1. For a particle with momentum ##k##, the two transverse polarization vectors ##\epsilon({\bf k}, \lambda_{1})## and ##\epsilon({\bf k}, \lambda_{1})## are purely spatial and orthogonal to ##\bf k##, that is,
    ##\epsilon^{0}({\bf k}, \lambda_{1}) = 0,##
    ##\epsilon({\bf k}, \lambda_{1})\cdot{k} = 0,##
    ##\epsilon^{0}({\bf k}, \lambda_{2}) = 0,##
    ##\epsilon({\bf k}, \lambda_{2})\cdot{k} = 0.##
    2. The third, longitudinal, polarization vector ##\epsilon({\bf k}, \lambda_{3})##}, for a particle with momentum ##k##, is timelike positive, orthogonal to ##k## as well as the transverse polarization vectors, and has unit negative norm, that is,
    ##\epsilon^{0}({\bf k}, \lambda_{3}) > 0,##
    ##\epsilon({\bf k}, \lambda_{3})\cdot{k} = 0,##
    ##\epsilon({\bf k}, \lambda_{3})\cdot{\epsilon({\bf k}}, \lambda_{1}) = 0,##
    ##\epsilon({\bf k}, \lambda_{3})\cdot{\epsilon({\bf k}}, \lambda_{2}) = 0,##
    ##\epsilon({\bf k}, \lambda_{3})\cdot{\epsilon({\bf k}}, \lambda_{3}) = -1.##

    3. We can infer from the orthogonality of the polarization vectors that the longitudinal polarization vector of a particle points in the direction of momentum of the particle.

    --------------------------------------------------------------------------------------------------------------------------------------

    Are these facts true for any gauge choice of the electromagnetic four-vector ##A^{\mu}##?
     
  2. jcsd
  3. Mar 20, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Mar 20, 2017 #3

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Do you mean a photon with momentum ##k##? Your notation and your question seem to imply that.
     
  5. Mar 23, 2017 #4
    There is a longitudinal polarization. This is for a W or Z boson.
     
  6. Mar 23, 2017 #5

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    But you refer to a gauge choice for the electromagnetic 4-vector. That's not the same as a W or Z boson. Which are you asking about?
     
  7. Mar 24, 2017 #6

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    In the electroweak standard model (as for any Higgsed gauge theory) the unitary gauge reveals the particle content explicitly. It's drawback is that the proper vertex functions are not explicitly renormalizable in this gauge, but to get the physical degrees of freedom, it's great. The great discovery by 't Hooft and Veltman in 1971 was thas one can choose a renormalizable gauge (called ##R_{\xi}## gauges), where the proper vertex functions are manifestly renormalizable since the power counting works as for scalar fields, i.e., the gauge-boson propogator falls qudratically with momentum in these gauges. Together with dim. reg. that enabled them to prove both the renormalizability of non-abelian Higgsed gauge theories and the unitarity and physicality of the S-matrix. It makes clear that both the Faddeev-Popov ghosts and the "would-be Goldstone bosons" conspire with the unphysical polarization degree of freedom of the gauge bosons to cancel these unphysical degrees out of the S-matrix (at any order in perturbation theory), and this makes the Standard Model a physically consistent QFT of the electroweak interaction.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Polarization of gauge bosons and gauge choice
  1. DoF of a gauge boson (Replies: 2)

  2. Gauge group choice (Replies: 1)

Loading...