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

I Again Gauge Transformations

  1. Apr 5, 2016 #1
    Due to its form, gauge transformations for the typical electrodynamics potentials are "local" in nature. That`s: they exists for path connected topological spaces. So, there exists global gauge transformations or are all of them local in nature?. If the answer is "yes", i.e. if there are global gauge transformations our ideas about electrodynamics are quite general, if the answer is "no", then we cannot have a global equivalence of the different gauges, and the results obtained in one gauge are not globally equivalent to the results in any other gauge. What do you think???.
     
  2. jcsd
  3. Apr 5, 2016 #2

    ShayanJ

    User Avatar
    Gold Member

    I'm not sure I understand your equation. But actually when people talk about a gauge theory, they mean some equations that already is symmetric under some global transformation and they're going to see what they can do to it to make it symmetric under the local version of that transformation.
     
  4. Apr 5, 2016 #3
    Well, let`s put the question in this form: any vector A with a zero curl ∇×A=0 can be expressed "locally" as a gradient A = ∇Ω for some function Ω. But its global version, .i.e. all along the topological space, is valid for path connected spaces only. For this reason the integral ∫A.dr is not zero for all closed trajectories when the space is not path connected. In the case of Maxwell equations the potentials are introduced solving two of the equations, hence the other two define the potentials. But the point is that that equations for the potentials admits an infinite dimensional symmetry that we call "gauge symmetry" , so the system is constrained in its Hamiltonian form and Noether second theorem tell us arbitrary functions exists in the solution. So, in this form the set of equations is underdetermined and no solution can be obtained; at least not in the sense of classical mathematical physics. To solve the equations we introduce a gauge condition to break the symmetry. To introduce such condition we use a gauge transformation that involves gradients -in its four dimensional form is a four gradient- so we suppose that a "gauge function" exists. E.g. when we transform from Coulomb to Lorenz gauge we suppose that a gauge function exists. J. D. Jackson believes that he knows how to obtain explicitly such a function. Here is where the question of global and local becomes critical: the gauge function exists locally -or that we believe and Jackson helps us to believe such a thing- but is the case that the gauge function exists globally???.
     
  5. Apr 5, 2016 #4

    ShayanJ

    User Avatar
    Gold Member

    At first I thought, by local and global, you mean the usual meaning but it seems you mean something else.
    I don't think you need a gauge function to fix a gauge, because you don't actually do a gauge transformation, you just assume your solution satisfies an equation.
     
  6. Apr 5, 2016 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    I wish I knew the whole mathematically complicated answer to your problem, by my college education in gauge theories was only algebraic, no differential geometry.
    I can only offer an article which should address your query (I hope): http://arxiv.org/abs/1211.6420v3
     
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: Again Gauge Transformations
  1. Gauge transformation (Replies: 9)

Loading...