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Vortices - differentiation of the field in abelian Higgs

  1. Apr 10, 2013 #1
    In Zee's "Nutshell QFT" (chapter V.7) or Shifman's book "Avanced Topics in QFT" (section 10) when they talk about vortices, they claim:

    if [itex]\phi(r,\theta)[/itex] goes to [itex]\nu[/itex]exp( i[itex]\theta[/itex]) as r goes to infinity

    then [itex]\partial_{i}[/itex][itex]\phi[/itex] becomes [itex]\nu[/itex](1/r)

    I do not see how? Is the phase equal to 1/r if r goes to r???

    thank you
    Last edited: Apr 10, 2013
  2. jcsd
  3. Apr 10, 2013 #2


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    Just express the gradient in polar coordinates:

    $$\nabla f =( \partial_r f )\hat{e}_r + \frac{1}{r} (\partial_\theta f )\hat{e}_\theta.$$
  4. Apr 10, 2013 #3
    thanks, fzero!
  5. Apr 11, 2013 #4
    Wait, still not clear...

    Especially, what Shifman does or Rubakov in "Classical Theory of Gauge Theories" on page 159, of which I attached a pdf copy down below. Where does the minus sign come from? Are the partial derivatives in polar or x-y coordinates? I know that the 1/r comes from the gradient somehow, as fzero pointed out, but why the Levi-Cita symbol?

    I'm sorry, scine I know that it must be super simple and "obvious", but I can't see it.

    Very grateful in advance for any extra hints!

    Attached Files:

  6. Apr 11, 2013 #5


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    You should work out ##\partial_i \theta## using the usual relation from polar coordinates that ##\theta = \tan^{-1}(y/x)##. The relative minus sign between the ##x## and ##y## components is related to the appearance of ##x/y## in this formula.
  7. Apr 12, 2013 #6
    Ok, after all not so super obvious:

    Given z = arctan(y/x) = arctan(yx⁻¹):

    ∂z/∂x = [1/(1 + (y/x)²)] * (∂/∂x)(yx⁻¹)
    = [1/(1 + (y/x)²)] * (-yx⁻²)
    = [1/(1 + y²/x²)] * (-y/x²)
    = -y / (x² + y²).

    ∂z/∂y = [1/(1 + (y/x)²)] * (∂/∂y)(yx⁻¹)
    = [1/(1 + y²/x²)] * (x⁻¹)
    = [1/(1 + y²/x²)] * (1/x)
    = 1/(x + y²/x)
    = x / (x² + y²).

    Arghhh, those textbooks that throw just equations at you without explaining anything.
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