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Gauge Fixing Term and Equations of Motion

  1. Jan 6, 2015 #1
    Lets take QED just to simplify. When we are doing Path Integrals and we want to "fix the Gauge":
    1 we add in the integral a delta(F) -meaning that we are going to integrate only in one representative of each class of equivalent configuration-
    2 We take some factors out because they are constant and we don´t need them in the Path Integral Approach
    3 We arrive to the same action than before but with a Gauge Fixing term "1/(2*epsilon)* (∂μAμ)2.

    So the lagrangian given by 3 is the same QED lagrangian up to a multiplicative factor that should not change anything. Nevertheless, if we derive the equations of motion of this new lagrangian, I think that we get a different set of equations.

    So, how can it be possible that we only get rid of some constant factor but we changed the path of the field configurations that makes the variation of the lagrangian equal to 0. What am I missing?

    As always, thanks in advance for your valuable help!!!!
     
  2. jcsd
  3. Jan 6, 2015 #2

    dextercioby

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    The 3 steps you mention are unknown to me, I'm only familiar to the BRST (Lagrangian/Batalin-Vilkovisky) gauge fixing method. The point of gauge fixing is indeed computing the path integral only with respect to 'true' field configurations which are counted only once, in the sense that gauge freedom should be somehow removed/replaced. This removal is always done with auxiliary fields, the ghost fields, whose role is to make sure a rigid symmetry (the BRST one) is replacing the unsatisfactory gauge one. In a sense, the gauge fixing is a continued process, for you can use the BRST symmetry to construct valid interacting theories (QED, Yang Mills from a collection of uncoupled real vector fields, QCD, perturbative gravity etc.), then use the same BRST you developed for the free non-interacting fields formalism to gauge fix the resulting coupled theory, so that the path integral (generating functional for the disconnected Green functions which helps you compute the S-matrix elements - especially for tree scattering processes) will be finite.

    The role of the eom is important only through the solution of the free fields eom, the latter being linear and thus with solutions relatively easy to obtain. You use from these solutions the amplitudes and the polarization/helicity vectors which then help you with the scattering amplitudes. The eom don't mean too much for the process of gauge fixing itself.
     
    Last edited: Jan 7, 2015
  4. Jan 6, 2015 #3
    Thanks dextercioby. Im not familiar with BRST math, I know that is the general method to deal with this issues in Non Abelian Gauges, nevertheless, I wanted to understand first QED Gauge issues and then generalize them. And in QED, the Gauge Fixing can be simplified following the three steps mentioned above.

    I know that eom are important, in practice, for free fields. I just thought about them in the interacting Lagrangian theory and I happen to find myself in a contradiction. How is it possible that the eom changes with the Lagrangian modified by 1), 2) and 3) of the first post, if all the changes in those steps only modified the Lagrangian in multiplicative factors???

    Surely Im understanding something wrong but I cant see.

    Again, thanks all the same!
     
  5. Jan 7, 2015 #4

    vanhees71

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  6. Jan 7, 2015 #5
    Thanks Vanhees, congratulations for your work. I will read it thoroughly. Can you please tell me where in your paper my question is answered or sort of.

    Ps: I took a light speed view of it and I saw, in chapter called (Canonical Path Integral), an approach not so traditional to Path Integrals where you sort of derive a Path Integral with no redundant degrees of freedoms (you dont integrate two of the four A fields). I was hoping that something like that should exist.
     
  7. Jan 7, 2015 #6

    vanhees71

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    It's chapter 6 for QED. You need the canonical path integral for all theories that contain (time-) derivative couplings, which is the case for scalar QED in many gauges. The point, not to use the radiation gauge for free fields is that you like to have a Lorentz-covariant formalism.
     
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