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Lagrangian invariant but Action is gauge invariant

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data

    So I'm having some difficulty with my QFT assignment. I have to solve the following problem.

    In three spacetime dimensions (two space plus one time) an antisymmetric Lorentz tensor
    F[itex]^{\mu\nu}[/itex] = -F[itex]^{\nu\mu}[/itex] is equivalent to an axial Lorentz vector, F[itex]^{\mu\nu}[/itex] = e[itex]^{\mu\nu\lambda}[/itex]F[itex]_{\lambda}[/itex]. Consequently, in 3D
    one can have a massive photon despite unbroken gauge invariance of the electromagnetic
    field A[itex]_{\mu}[/itex]. Indeed, consider the following Lagrangian:

    L = -(1/2)*F[itex]_{\lambda}[/itex]F[itex]^{\lambda}[/itex] + (m/2)*F[itex]_{\lambda}[/itex]A[itex]^{\lambda}[/itex] (6)

    where

    F[itex]_{\lambda}[/itex](x) = (1/2)*[itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex]F[itex]^{\mu\nu}[/itex] = [itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex][itex]\partial[/itex][itex]^{\mu}[/itex]A[itex]^{\nu}[/itex],

    or in components, F[itex]_{0}[/itex] = -B, F1 = +E[itex]^{2}[/itex], F[itex]_{2}[/itex] = -E[itex]^{1}[/itex].

    (a) Show that the action S = [itex]\int[/itex]d[itex]^{3}[/itex]x*L is gauge invariant (although the Lagrangian (6) is not invariant).



    So I tried substituting A[itex]^{\lambda}[/itex] -> A[itex]^{\lambda'}[/itex] = A[itex]^{\lambda}[/itex] + [itex]\partial[/itex][itex]^{\lambda}[/itex][itex]\Lambda[/itex]
    and F[itex]^{\lambda}[/itex] -> F[itex]^{\lambda'}[/itex] = [itex]\epsilon[/itex][itex]^{\lambda\mu\nu}[/itex][itex]\partial[/itex][itex]_{\mu}[/itex]A[itex]_{\nu}[/itex]'

    then I obtained L' = L + (1/2)*[ F[itex]_{\lambda}[/itex] [itex]\epsilon^{\lambda\mu\nu}[/itex][itex]\partial[/itex][itex]_{\mu}[/itex] [itex]\partial[/itex][itex]_{\nu}[/itex] [itex]\Lambda[/itex] + some other terms]

    What I don't understand is how these leftover terms would vanish after being integrated (to obtain S'), but they don't all vanish if they are not integrated (since L is not invariant). Is there some kind of special mathematical trick I have to use? I just don't see how I can integrate terms like [itex]\int[/itex]d[itex]^{3}[/itex]x F[itex]_{\lambda}[/itex][itex]\epsilon[/itex][itex]^{\lambda\mu\nu}[/itex][itex]\partial[/itex][itex]_{\mu}[/itex][itex]\partial[/itex][itex]_{\nu}[/itex][itex]\Lambda[/itex]
     
  2. jcsd
  3. Dec 29, 2011 #2
    Well, I can see how Fλ ϵλμν∂μ ∂ν Λ would vanish. The \mu and \nu are symmetric wrt to exchange for the partial derivatives, but the indices of the \epsilon symbol are totally antisymmetric. Multiplying a symmetric tensor and an antisymmetry one gives zero always.
     
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